Dani and Mike are two kids ,They are playing games all day and when they don't find a game to play they invent a game . There is about an hour to arrive to school, because they love playing with strings Dani invented a game , Given a string and the winner is the first who form a palindrome string using all letters of this string according to the following sample rules : 1- player can rearrange letters to form a string . 2- the formed string must be palindrome and use all letters of the given string. 3- if there is more than one string chose the lexicographically smallest string . EX: string is "abacb" player can form : "abcba" and "bacab" ,but "abcba" is the lexicographically smallest. Mike asked you to write a Program to compute the palindrome string so he can beat Dani.
Input
Your program will be tested on one or more test cases. The first line of the input will be a single integer T, the number of test cases (1 ≤ T ≤ 1000). Every test case on one line contain one string ,the length of the string will not exceed 1000 lower case English letter.
Output
For each test case print a single line containing the lexicographically smallest palindrome string according to the rules above. If there is no such string print "impossible"
Examples
Input
4 abacb acmicpc aabaab bsbttxs
Output
abcba impossible aabbaa bstxtsb
Note
Palindrome string is a string which reads the same backward or forward.
Lexicographic order means that the strings are arranged in the way as they appear in a dictionary.
AC代码（没有想到aabba这种情况的输出应该是aabab）
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main()
{
char a[1000+10];
int t, i, b[1000+10], j;
scanf("%d",&t);
getchar();
while(t--)
{
int f = 0;
char g;
int g1;
memset(b, 0, sizeof(b));
cin>>a;
int n = strlen(a);
for(i = 0; i<n; i++)
{
b[a[i]-'a']++;
}
for(i = 0; i<26; i++)
{
if(b[i]%2!=0)
{
f++;
g = i+'a';
g1 = b[i];
b[i]--;
}
}
if(f>1)
printf("impossible\n");
else
{
for(i = 0; i<26; i++)
{
for(j = 0; j<b[i]/2; j++)
printf("%c",i+'a');
}
if(f!=0)
{
printf("%c",g);
}
for(i = 25; i>=0; i--)
{
for(j = 0; j<b[i]/2; j++)
printf("%c",i+'a');
}
printf("\n");
}
}
return 0;
}