Marina and Vasya

 Marina and Vasya

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Marina loves strings of the same length and Vasya loves when there is a third string, different from them in exactly t characters. Help Vasya find at least one such string.

More formally, you are given two strings s1, s2 of length n and number t. Let's denote as f(a, b) the number of characters in which strings a and b are different. Then your task will be to find any string s3 of length n, such that f(s1, s3) = f(s2, s3) = t. If there is no such string, print  - 1.

Input

The first line contains two integers n and t (1 ≤ n ≤ 105, 0 ≤ t ≤ n).

The second line contains string s1 of length n, consisting of lowercase English letters.

The third line contain string s2 of length n, consisting of lowercase English letters.

Output

Print a string of length n, differing from string s1 and from s2 in exactly t characters. Your string should consist only from lowercase English letters. If such string doesn't exist, print -1.

Sample test(s)

input

3 2
abc
xyc

output

ayd

input

1 0
c
b

output

-1

 

 建设性设计问题：

　　　　创造不同即是创造相同：t = n - t;

　　　　首先考虑ch1和ch2相同的字母，如果这样的字母能够满足要求，则结束；

　　　　如果还不够，再考虑ch1和ch2中不同的字母，不同ans即为一个像ch1，一个像ch2；

　　　　如果还不够，则无解；

　　　　想要相同很简单，赋值即可，想要不同就拿'a','b','c'去比较ch1和ch2的字母，总有都不一样的，赋值也即可；

　　　　如上所说，开始先记录相同数目x，和不同数目y，x+y/2即为最大相同值，若t>x+y/2显然无解；

　　　　若有解，先判断相同数目x能否满足要求，tt = t - x;若tt <= 0 则表示满足，tt = 0;反而在相同字母都兑换的情况下t = x，搞不同的字母，ch1一个(sign = 1) and ch2一个(sign = 0),同时tt--;就ok了；

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
const int maxn = 1e5+5;
using namespace std;
char ch1[maxn];
char ch2[maxn];
char ans[maxn];
bool flag[maxn];
int main(){
    int n,t;
    int x = 0,y = 0;
    bool sign = 0;
    scanf("%d%d",&n,&t);
    t = n - t;
    scanf("%s%s",ch1,ch2);
    for(int i = 0; i < n; i++){
        if(ch1[i] == ch2[i]){
            flag[i] = 1;
            x++;
        }
        else y++;
    }
    if(t > x+y/2){
        puts("-1");
        return 0;
    }
    else {
        int tt = t - x;
        if(tt < 0)tt= 0;
        else t = x;
        for(int i = 0; i < n; i++){
            if(flag[i] && t){
                ans[i] = ch1[i];
                t--;
            }
            else if(tt||sign){
                if(sign){
                    sign = 0;
                    ans[i] = ch2[i];
                }
                else {
                    tt--;
                    ans[i] = ch1[i];
                    sign = 1;
                }
            }
            else {
                for(int j = 0; j < 3; j++){
                    if(ch1[i] != 'a'+j && ch2[i] != 'a'+j){
                        ans[i] = 'a'+j;
                        break;
                    }
                }//！！！有选择恐惧症的我一直在纠结不同的话该放什么字母上去，这是姿势。。
            }
        }
    }
    ans[n] = '\0';
    cout<< ans<< endl;
    return 0;
}