### ABABAABA

#### Description:

You are given a uniformly randomly generated string S, consisting of letters from the set {""A"", ""B""}. Your task is to find a string T that appears in S as a subsequence exactly twice. In other words, you need to find such a string T, that there exist exactly two sets of indexes i1, i2, ..., i|T| and j1, j2, ..., j|T| such that there exists some k, where ik jk and S{i1...i|T|} = S{j1...j|T|} = T. Input The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The first and only line of each test case contains a single string S. The string S was generated randomly. For a generating string S, we first choose an integer N denoting a length of S. After that every symbol of the string S is chosen randomly from the set {""A"", ""B""} and the both symbols have equal probability to be chosen. Note that N is not choosen randomly. Output For each test case, output a string that occurs exactly twice as a subsequence in S, or output -1 if there is no such string. If there are more than one possible subsequences occurring exactly two times, you can print any one of them. Constraints 1 <=T<= 10 Explanation Test case #1: The string ""AAAA"" appears once as a subsequence in itself. The string ""AAA"" appears four times as a subsequence in ""AAAA""; possible positions: {2, 3, 4}, {1, 3, 4}, {1, 2, 4}, {1, 2, 3}. The strings ""AA"" and ""A"" also appear in ""AAAA"" as a subsequence strictly more than twice. So, there is no string of ""AAAA"", which appears exactly twice. Hence answer is -1. Test case #2: Two occurrences of ""B"" in ""BAB"" are {1} and {3} (1-based indexing)." Test Case 1 Input (stdin) 2 AAAA BAB Expected Output -1 B Test Case 2 Input (stdin) 3 AAAAA BABBA AAA Expected Output -1 A -1

#### Program :

#include <stdlib.h>

#include<stdio.h>

#include<string.h>

#include<math.h>

int main()

{int t;

scanf("%d",&t);

while(t--)

{char ch;

scanf("%s",ch);

int i,j,x=0,y=0;

for(i=0;ch[i]!='\0';i++)

{if(ch[i]=='A')

x++;

else

y++;

}

if(x==2)

{printf("A\n");continue;

}

if(y==2)

{printf("B\n");continue;

}

if(x==1 || y==1)

{printf("-1\n");continue;

}

for(j=0;j<i;j++)

{int ans=0;

while(ch[j]!='A' && j<i)

{ans++;

j++;

}

if(ans==2)

{for(x=0;x<j-2;x++)

{if(ch[x]=='A')

printf("%c",ch[x]);

}

printf("B");

for(x=j;x<i;x++)

{if(ch[x]=='A')

printf("%c",ch[x]);

}

printf("\n");

j=-1;

}

if(j==-1)

break;

}

if(j==-1)

{continue;

}

for(j=0;j<i;j++)

{int ans=0;

while(ch[j]!='B' && j<i)

{ans++;

j++;

}

if(ans==2)

{for(x=0;x<j-2;x++)

{if(ch[x]=='B')

printf("%c",ch[x]);

}

printf("A");

for(x=j;x<i;x++)

{if(ch[x]=='B')

printf("%c",ch[x]);

}

printf("\n");

j=-1;

}

if(j==-1)

break;

}

if(j==-1)

{continue;

}

printf("-1\n");

}

return 0;

}