傳送門

A. Vasya And Password

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya came up with a password to register for EatForces — a string ss. The password in EatForces should be a string, consisting of lowercase and uppercase Latin letters and digits.

But since EatForces takes care of the security of its users, user passwords must contain at least one digit, at least one uppercase Latin letter and at least one lowercase Latin letter. For example, the passwords "abaCABA12", "Z7q" and "3R24m" are valid, and the passwords "qwerty", "qwerty12345" and "Password" are not.

A substring of string ss is a string x=slsl+1…sl+len−1(1≤l≤|s|,0≤len≤|s|−l+1)x=slsl+1…sl+len−1(1≤l≤|s|,0≤len≤|s|−l+1). lenlen is the length of the substring. Note that the empty string is also considered a substring of ss, it has the length 00.

Vasya's password, however, may come too weak for the security settings of EatForces. He likes his password, so he wants to replace some its substring with another string of the same length in order to satisfy the above conditions. This operation should be performed exactly once, and the chosen string should have the minimal possible length.

Note that the length of ss should not change after the replacement of the substring, and the string itself should contain only lowercase and uppercase Latin letters and digits.

Input

The first line contains a single integer TT (1≤T≤1001≤T≤100) — the number of testcases.

Each of the next TT lines contains the initial password s (3≤|s|≤100)s (3≤|s|≤100), consisting of lowercase and uppercase Latin letters and digits.

Only T=1T=1 is allowed for hacks.

Output

For each testcase print a renewed password, which corresponds to given conditions.

The length of the replaced substring is calculated as following: write down all the changed positions. If there are none, then the length is 00. Otherwise the length is the difference between the first and the last changed position plus one. For example, the length of the changed substring between the passwords "abcdef" →→ "a7cdEf" is 44, because the changed positions are 22 and 55, thus (5−2)+1=4(5−2)+1=4.

It is guaranteed that such a password always exists.

If there are several suitable passwords — output any of them.

Example

input

Copy

2
abcDCE
htQw27

output

Copy

abcD4E
htQw27

Note

In the first example Vasya's password lacks a digit, he replaces substring "C" with "4" and gets password "abcD4E". That means, he changed the substring of length 1.

In the second example Vasya's password is ok from the beginning, and nothing has to be changed. That is the same as replacing the empty substring with another empty substring (length 0).

題意：

• 最少的操作使原串變成有數字+大小寫字母的串。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
	int w;
	cin>>w;
	while(w--)
	{
		int n=0,m=0,t=0;
		string s;
		cin>>s;
		for(int i=0;i<s.size();i++)
		{
			if(s[i]>='a'&&s[i]<='z')
			{
				n++;
			}
			if(s[i]>='A'&&s[i]<='Z')
			{
				m++;
			}
			if(s[i]>='0'&&s[i]<='9')
			{
				t++;
			}
		}
		for(int i=0;i<sizeof(s);i++)
		{
			if(t&&n&&m)
			break;
			if(s[i]>='a'&&s[i]<='z'&&n>1)
			{
				if(m==0)
				{
					m++;
					s[i]='A';
					continue;
				}
				if(t==0)
				{
					t++;
					s[i]='1';
					continue;
				}
			}
			if(s[i]>='A'&&s[i]<='Z'&&m>1)
			{
				if(n==0)
				{
					n++;
					s[i]='a';
					continue;
				}
				if(t==0)
				{
					t++;
					s[i]='1';
					continue;
				}
			}
			if(s[i]>='0'&&s[i]<='9'&&t>1)
			{
				if(n==0)
				{
					n++;
					s[i]='a';
					continue;
				}
				if(m==0)
				{
					m++;
					s[i]='A';
					continue;
				}
			}
		}
		cout<<s<<endl;
	}
	return 0;
} 

B. Relatively Prime Pairs

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a set of all integers from ll to rr inclusive, l<rl<r, (r−l+1)≤3⋅105(r−l+1)≤3⋅105 and (r−l)(r−l) is always odd.

You want to split these numbers into exactly r−l+12r−l+12 pairs in such a way that for each pair (i,j)(i,j) the greatest common divisor of ii and jj is equal to 11. Each number should appear in exactly one of the pairs.

Print the resulting pairs or output that no solution exists. If there are multiple solutions, print any of them.

Input

The only line contains two integers ll and rr (1≤l<r≤10181≤l<r≤1018, r−l+1≤3⋅105r−l+1≤3⋅105, (r−l)(r−l) is odd).

Output

If any solution exists, print "YES" in the first line. Each of the next r−l+12r−l+12 lines should contain some pair of integers. GCD of numbers in each pair should be equal to 11. All (r−l+1)(r−l+1) numbers should be pairwise distinct and should have values from ll to rr inclusive.

If there are multiple solutions, print any of them.

If there exists no solution, print "NO".

Example

input

Copy

1 8

output

Copy

YES
2 7
4 1
3 8
6 5

題意：

• 給出一組從l到r的所有整數，（r-l）總是奇數。
• 將這些數字精確地分成(r-l+1)/2對，使得對於每對i、j的最大公約數等於1。
• 列印生成的對。如果有多個解決方案，請列印其中任何一個。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#define ll long long
using namespace std;
int main()
{
	ll l,r;
	cin>>l>>r;
	cout<<"YES"<<endl;
	for(ll i=l;i+1<=r;i+=2)
	{
		cout<<i<<" "<<i+1<<endl;
	 } 
	return 0;
}

C. Vasya and Multisets

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has a multiset ss consisting of nn integer numbers. Vasya calls some number xx nice if it appears in the multiset exactly once. For example, multiset {1,1,2,3,3,3,4}{1,1,2,3,3,3,4} contains nice numbers 22 and 44.

Vasya wants to split multiset ss into two multisets aa and bb (one of which may be empty) in such a way that the quantity of nice numbers in multiset aa would be the same as the quantity of nice numbers in multiset bb (the quantity of numbers to appear exactly once in multiset aaand the quantity of numbers to appear exactly once in multiset bb).

Input

The first line contains a single integer n (2≤n≤100)n (2≤n≤100).

The second line contains nn integers s1,s2,…sn (1≤si≤100)s1,s2,…sn (1≤si≤100) — the multiset ss.

Output

If there exists no split of ss to satisfy the given requirements, then print "NO" in the first line.

Otherwise print "YES" in the first line.

The second line should contain a string, consisting of nn characters. ii-th character should be equal to 'A' if the ii-th element of multiset ssgoes to multiset aa and 'B' if if the ii-th element of multiset ss goes to multiset bb. Elements are numbered from 11 to nn in the order they are given in the input.

If there exist multiple solutions, then print any of them.

Examples

input

Copy

4
3 5 7 1

output

Copy

YES
BABA

input

Copy

3
3 5 1

output

Copy

NO

題意：

• 定義 nice numbers為一組數裡只出現一次的數
• 給你n個數讓你把這組數分成兩個陣列 A,B 若這兩個陣列nice numbers數相等 輸出Yes並且指出每個數在A還是B裡 如果有多種情況 輸出其中一種情況即可

題解： 

• 若這組數裡面只出現一次的數一共有偶數個說明可以組成

• 若這組數裡面只出現一次的數一共有奇數個 並且出現次數大於三的個數為0 說明不可以組成 

• 這組數裡面只出現一次的數一共有奇數個 並且出現次數大於三的個數不為0 說明可以組成 比如 出現了三次5 可以將一個5放在一個數組 其他的放在另一個數組 這樣就可以看成出現次數為1的個數為偶數了 

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring> 
#include<map>
#define N 0x3f3f3f3f
#define ll long long
using namespace std;
int main()
{
	int n;
	cin>>n;
	int a[1000];
	int f[1000];
	int t[1000];
	for(int i=1;i<=n;i++)
	{
		cin>>t[i];
		a[t[i]]++;
	}
	int x=0,y=0;
	for(int i=1;i<=100;i++)
	{
		if(a[i]==1)
		{
			x++;
		}
		if(a[i]>2)
		y++;
	}
	if(x%2==1&&y==0)
	{
		cout<<"NO"<<endl;
	}
	else
	{
		cout<<"YES"<<endl;
		int j=1;
		for(int i=1;i<=x/2;j++)
		{
			if(a[t[j]]==1)
			{
				f[j]=1;
				i++;
			}
		}
		if(x%2)
		{
			for(int i=1;i<=n;i++)
			{
				if(a[t[i]]>2)
				{
					f[i]=1;
					break;
				}
			}
		}
		for(int i=1;i<=n;i++)
		{
			if(f[i]==1)
			cout<<'A';
			else
			cout<<'B';
		}
	}
	return 0;
}