Codeforces Round #303 Equidistant String

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 andt of the same length consisting of digits zero and one as the number of positionsi, such that s_i isn't equal tot_i.

As besides everything else Susie loves symmetry, she wants to find for two stringss andt of lengthn such stringp of lengthn, that the distance fromp to s was equal to the distance fromp tot.

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 lengthn.

The second line contains string t of lengthn.

The length of string n is within range from1 to10⁵. 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.

Sample test(s)

Input

00011011

Output

0011

Input

000111

Output

impossible

#include <iostream>#include <cstring>#include <stack>#include <queue>#include <algorithm>#include <cmath>#include <cstdlib>#define LL long long#define INF 0x3f3f3f3fusing namespace std;char s[100010],a[1000010],p[100010];int main(){    while(cin>>s>>a)    {        int len = strlen(s);        int ant = 0;        for(int i=0;i<len;i++)        {            if(s[i] == a[i])                p[i] = a[i];            else            {                ant++;                if(ant%2)                    p[i] = s[i];                else                    p[i] = a[i];            }        }        if(ant % 2)            cout<<"impossible"<<endl;        else            cout<<p<<endl;    }    return 0;}