Codeforces Round #303 (Div.2)-B. Equidistant String（模拟）

B. Equidistant String

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

00011011

output

0011

input

000111

output

impossible

Note

In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.

题意：

给出s，t两个n长的字符串。要你找出是否存在一个字符串p使得，|s-p|==|t-p|。

思路：

只要使不相同的s【i】，他【i】存在偶数个，那么就从s取一半，t里取一半就能构成所需字符串了。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<cstdio>#include<vector>using namespace std;int main(){#ifdef zscfreopen("input.txt","r",stdin);#endifstring s1,s2;int i,j,k,cnt;while(cin >> s1 >> s2){string ve;cnt = 0;for(i=0;s1[i];++i)cnt+=(s1[i]!=s2[i]);if(cnt&1){printf("impossible\n");}else {int c=0;for(i=0;s1[i];++i){if(s1[i]==s2[i])ve+=s1[i];else {if(c<cnt/2)ve+=s1[i],c++;else ve+=s2[i];}}cout << ve << endl;}}return 0;}