B. Valued Keys

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本文介绍了一个名为f的神秘函数，该函数接受两个相同长度的字符串，并返回一个新的字符串，其中每个字符都是输入字符串相应位置上的最小值。文章探讨了如何找到一个字符串z，使得函数f应用于给定字符串x和z时，结果等于另一个给定字符串y。

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", and f("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.

Examples

input
ab
aa

output
ba

input
nzwzl
niwel

output
xiyez

input
ab
ba

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".

题解：根据题意，设三个字符串。然后根据函数特征扫一遍逐个判断，当x[i]>y[i]时，z+=y[i].当x[i]==y[i]，z+=x[i]+temp（注意，x[i]+temp必须不大于z）

当x[I]<y[I]时，直接输出-1退出。

代码如下：

//
//  main.cpp
//  B. Valued Keys
//
//  Created by 徐智豪 on 2017/4/16.
//  Copyright © 2017年 徐智豪. All rights reserved.
//

#include <iostream>
#include <string>
using namespace std;
string x,y,z;
int main(int argc, const char * argv[]) {
    cin>>x>>y;
    int flag=1;
    int lengthx=(int)x.size();
    int lengthy=(int)y.size();
    if(lengthx==lengthy)
    for(int i=0;i<lengthx;i++)
    {
        if(x[i]==y[i])
        {
            if(x[i]<'z')
            z+=(x[i]+1);
            else
                z+=x[i];
        }
        else if(x[i]>y[i])
        {
            z+=y[i];
        }
        else if(x[i]<y[i])
        {
            flag=0;
            break;
        }
    }
    if(flag)
    cout<<z<<endl;
    else
        cout<<-1<<endl;
    return 0;
}