Description

Little C loves number «3» very much. He loves all things about it.

Now he has a positive integer nn. He wants to split nn into 33 positive integers a,b,ca,b,c, such that a+b+c=na+b+c=n and none of the 33 integers is a multiple of 33. Help him to find a solution.

Input

A single line containing one integer nn (3≤n≤1093≤n≤109) — the integer Little C has.

Output

Print 33 positive integers a,b,ca,b,c in a single line, such that a+b+c=na+b+c=n and none of them is a multiple of 33.

It can be proved that there is at least one solution. If there are multiple solutions, print any of them.

Sample Input

Input

3

Output

1 1 1

Input

233

Output

77 77 79

程式碼如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#define max(a,b)   (a>b?a:b)
#define min(a,b)   (a<b?a:b)
#define swap(a,b)  (a=a+b,b=a-b,a=a-b)
#define maxn 320007
#define N 100000000
#define INF 0x3f3f3f3f
#define mod 1000000009
#define e  2.718281828459045
#define eps 1.0e18
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define ll long long
//std::ios::sync_with_stdio(false);
//cin.tie(NULL);
using namespace std;


int main()
{
    int n;
    cin>>n;
    if((n-2)%3==0)
        cout<<"1 2 "<<n-3<<endl;
    else
        cout<<"1 1 "<<n-2<<endl;
    return 0;
}
 

做的時候短路了，用了一個特別麻煩的方法，以此借鑑：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#define max(a,b)   (a>b?a:b)
#define min(a,b)   (a<b?a:b)
#define swap(a,b)  (a=a+b,b=a-b,a=a-b)
#define maxn 320007
#define N 100000000
#define INF 0x3f3f3f3f
#define mod 1000000009
#define e  2.718281828459045
#define eps 1.0e18
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define ll long long
//std::ios::sync_with_stdio(false);
//cin.tie(NULL);
using namespace std;



int main()
{
    int n;
    cin>>n;
    if(n==5)
        cout<<"1 2 2"<<endl;
    else if(n%3==0)
    {
        int m=n/3;
        if(m%3!=0)
            cout<<m<<" "<<m<<" "<<m<<endl;
        else
            cout<<m-1<<" "<<m-1<<" "<<m+2<<endl;
    }
    else
    {
        int m=n/3;
        int k=n%3;
        int a[3]={m,m,m+k};
        if(a[0]%3!=0&&a[1]%3!=0&&a[2]%3!=0)
            cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<endl;
        else
        {
            if(m%3==0)
            {
                a[0]--;
                a[1]--;
                a[2]+=2;
                if(a[0]%3!=0&&a[1]%3!=0&&a[2]%3!=0)
                    cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<endl;
                else
                    cout<<a[0]-1<<" "<<a[1]<<" "<<a[2]+1<<endl;
            }
            else
            {
                a[0]--;
                a[1]--;
                a[2]+=2;
                if(a[0]%3!=0)
                    cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<endl;
                else
                    cout<<a[0]-1<<" "<<a[1]+1<<" "<<a[2]<<endl;
            }
        }
    }
    return 0;
}