Find the Lost Sock （异或算法）

Alice bought a lot of pairs of socks yesterday. But when she went home, she found that she has lost one of them. Each sock has a name which contains exactly 7 charaters.

Alice wants to know which sock she has lost. Maybe you can help her.

Input

There are multiple cases. The first line containing an integer n (1 <=n <= 1000000) indicates that Alice bought n pairs of socks. For the following 2*n-1 lines, each line is a string with 7 charaters indicating the name of the socks that Alice took back.

Output

The name of the lost sock.

Sample Input

2
aabcdef
bzyxwvu
bzyxwvu
4
aqwerty
eas fgh
aqwerty
easdfgh
easdfgh
aqwerty
aqwerty
2
0x0abcd
0ABCDEF
0x0abcd

Sample Output

aabcdef
eas fgh
0ABCDEF

运用异或运算的一个性质。
X^X=0；相同数字异或答案为0；
所以最后的情况变成A^B^B^C^C^D^D=A^0^0^0=A

把一串分成单个的整数进行异或运算，结果很快就出来了

#include<stdio.h>
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
         getchar();                     
         int i,j;
         int a[8]={0};
         char s[10];
         for(i=0;i<2*n-1;i++)
         {
              gets(s);
              for(j=0;j<7;j++)
              a[j]=a[j]^s[j];
         }
         for(i=0;i<7;i++)
         printf("%c\n",a[i]);
    }
}