Problem: 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.

Examples

Input

3 2
abc
xyc


Output

ayd

Input

1 0
c
b


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100004;
char s[N], t[N], str[N];
int main() {
  int n, tr;
  scanf("%d %d", &n, &tr);
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (s[i] != t[i]) cnt++;
  if (tr < cnt) {
    int val = cnt / 2 + cnt % 2;
    if (tr < val)
      printf("-1\n");
    else {
      val = cnt - tr;
      val = val * 2;
      cnt = 1;
      for (int i = 1; i <= n; i++) {
        if (s[i] == t[i])
          str[i] = s[i];
        else {
          if (cnt <= val) {
            if (cnt % 2 == 1)
              str[i] = t[i];
            else
              str[i] = s[i];
            cnt++;
          } else {
            for (int j = 0; j < 26; j++) {
              if (((s[i] - 'a') != j) && ((t[i] - 'a') != j)) {
                str[i] = j + 'a';
                break;
              }
            }
          }
        }
      }
      printf("%s\n", str + 1);
    }
  } else {
    int val = tr - cnt;
    for (int i = 1; i <= n; i++) {
      if (s[i] == t[i]) {
        if (val > 0) {
          str[i] = ((s[i] - 'a') + 1) % 26 + 'a';
          val--;
        } else
          str[i] = s[i];
      } else {
        for (int j = 0; j < 26; j++) {
          if ((j != s[i] - 'a') && (j != t[i] - 'a')) {
            str[i] = j + 'a';
            break;
          }
        }
      }
    }
    printf("%s\n", str + 1);
  }
  return 0;
}