Description
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:

We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn’t equal to ti.

As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.

It’s time for Susie to go to bed, help her find such string p or state that it is impossible.

Input
The first line contains string s of length n.

The second line contains string t of length n.

The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.

Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line “impossible” (without the quotes).

If there are multiple possible answers, print any of them.

Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
题意：

给定两条长度相等的字符串s和t，字符串之间的距离为两串相同位置不相等元素的个数，问能否得到一个长度与两串相同的字符串p使得p的距离离s，t相等，若有遍输出p。

水啊！s和t的距离ans为偶数就可以，否则就没有。输出的话只要将s的前ans/2个不同的元素转换为与t相同即可得字符串p。
C语言版本一

#include<stdio.h>
#include<string.h>
#define N 100010
 
char s1[N], s2[N];
 
int main()
{
    int len;
    while(~scanf("%s%s", s1, s2))
    {
        int k=2, t=0;
        len=strlen(s1);
        for(int i=0;i<len;i++)
        {
            if(s1[i]!=s2[i])
            {
                t++;
            }
        }
        if(t%2!=0)
        {
            printf("impossible\n");
            continue;
        }
        for(int i=0;i<len;i++)
        {
            if(s1[i]==s2[i])
                printf("0");
            else
            {
                if(k%2)
                    printf("%c", s1[i]);
                else
                    printf("%c", s2[i]);
                k++;
            }
        }
        printf("\n");
    }
    return 0;
}

C语言版本二

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char *argv[]) {
	char s[100010],p[100010],t[100010];
	scanf("%s%s",&s,&p);
	int count=0;
	int i;
	strcpy(t,s);
	for(i=0;i<strlen(s);i++){
		if(s[i]!=p[i])
		{
			count++;			
		}
	}
	if(count%2==0){
		count/=2;
		for(i=0;i<strlen(s);i++){
            if(s[i]!=p[i]&&count){
                if(t[i]=='1')
                t[i]='0';
                else
                t[i]='1';
                count--;
            }
        }
        
		printf("%s\n",t);
		
	}
	
	else printf("impossible\n");
	return 0;
}

C++版本一

#include<iostream>
#include<cstring>
using namespace std;

string s,t,p;

int main(){
    cin>>s;
    cin>>t;
    int ans=0;
    p=s;
    int len=s.size();
    for(int i=0;i<len;i++){
        if(s[i]!=t[i]){
            ans++;
            t[i]=' ';    
        }
    }
    int temp=ans/2;
    if(ans%2==0){
        for(int i=0;i<len;i++){
            if(t[i]==' '&&temp){
                if(p[i]=='1')
                p[i]='0';
                else
                p[i]='1';
                temp--;
            }
        }
        for(int i=0;i<len;i++){
            cout<<p[i];
        }
    }
    else{
        cout<<"impossible";
    }
    return 0;
}