A positive integer is called a palindrome if its representation in the decimal system is the same when read from left to right and from right to left. For a given positive integer K of not more than 1000000 digits, write the value of the smallest palindrome larger than K to output. Numbers are always displayed without leading zeros.

Input

The first line contains integer t, the number of test cases. Integers K are given in the next t lines.

Output

For each K, output the smallest palindrome larger than K.

Example

Input:
2
808
2133

Output:
818
2222

Warning: large Input/Output data, be careful with certain languages

題意：輸出比X大的第一個回文字符串。

思路：先把X按左半邊為標準變成一個回文串X2，如果X2大於X，則輸出X2。 否則變大X2 ：

如果X2全部為9，則需要加一位，變為首尾為‘1’，之間為‘0’的回文串。

否則，從之間開始找第一位非‘9’的位置，自加1。然後中間取余變為‘0’。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include
 
<algorithm>
using namespace std;
const int maxn=1000010;
char c[maxn],c2[maxn];
int T,N,Len,a[maxn];
bool check9()
{
    for(int i=1;i<=Len;i++)
     if(c[i]!=‘9‘) return false;
    return true;
}
void Tochange()
{
    for(int i=1;i<=Len/2;i++) c2[i]=c[i];
    for(int i=Len/2+1;i<=Len;i++) c2[i]=c[Len+1-i];
}
bool Toupper()
{
    for(int i=1;i<=Len;i++) 
        if(c2[i]>c[i]) return true;
        else if(c2[i]<c[i]) return false; 
    return false;
}
int main()
{
    int i,j;
    scanf("%d",&T);
    while(T--){
        scanf("%s",c+1);
        Len=strlen(c+1);
        Tochange();
        if(Toupper()) {
            for(i=1;i<=Len;i++) putchar(c2[i]);
            cout<<endl;
            continue;
        }
        if(check9()) {
            putchar(‘1‘);
            for(i=1;i<Len;i++) putchar(‘0‘);
            putchar(‘1‘);
            cout<<endl;
            continue;
        }
         int np,Mid;
        if(Len&1) Mid=(Len+1)/2;
        else Mid=Len/2;
        for(np=Mid;np>=1;np--) if(c[np]!=‘9‘) break;
          c[np]++;
        for(i=np+1;i<=Mid;i++) c[i]=‘0‘;
        for(i=1;i<=Mid;i++) putchar(c[i]);
        for(i=Len/2;i>=1;i--) putchar(c[i]);
        cout<<endl;        
    }
    return 0;
}

SPOJ:The Next Palindrome(思維)