0519 G2n#W2B-D Valued Keys

摘要：给出两个字符串构造第三个串的方法，反过来求给出结果和其中一个求另一个。

原题目：CodeForces - 801B

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Valued Keys

You found a mysterious function f. The function takes two strings s₁ and s₂. These strings must consist only of lowercase English letters, and must be the same length.

The output of the function f is another string of the same length. The i-th character of the output is equal to the minimum of the i-th character of s₁ and the i-th character of s₂.

For example, f("ab", "ba") = "aa", andf("nzwzl", "zizez") = "niwel".

You found two strings x and y of the same length and consisting of only lowercase English letters. Find any string z such that f(x, z) = y, or print -1 if no such string z exists.

Input

The first line of input contains the string x.

The second line of input contains the string y.

Both x and y consist only of lowercase English letters, x and y have same length and this length is between 1 and 100.

Output

If there is no string z such that f(x, z) = y, print -1.

Otherwise, print a string z such that f(x, z) = y. If there are multiple possible answers, print any of them. The string z should be the same length as x and y and consist only of lowercase English letters.

Example

Input

abaa

Output

ba

Input

nzwzlniwel

Output

xiyez

Input

abba

Output

-1

Note

The first case is from the statement.

Another solution for the second case is "zizez"

There is no solution for the third case. That is, there is no z such that f("ab", z) =  "ba".

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题目理解：找出由z=f(x,y)得到的y=g(x,z)的表达式即可。

注意 ：不存在是输出 -1

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char X[102];
char Z[102];
char Y[102];
int len;
int main(){
scanf("%s%s",X,Z);
len=strlen(X);
bool ans = true;
for(int i=0;i<len;i++){
if(X[i]==Z[i]) Y[i]=Z[i];
else if(X[i]>Z[i]) Y[i]=Z[i];
else {
ans = false;
break;
}
}
if(ans) printf("%s",Y);
else printf("-1");
}

2017 05 19