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>
char s1[100100], s2[100100], s3[100100];
bool dist[100100];
int pos[100100];
int main() {
  int n, t, i;
  int num = 0, po = 0;
  memset(dist, false, sizeof(dist));
  scanf("%d%d", &n, &t);
  scanf("%s%s", s1, s2);
  for (i = 0; i < n; i++) s3[i] = '1';
  for (i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      dist[i] = true;
      pos[po++] = i;
      num += 2;
    } else
      num++;
  }
  if (num < 2 * (n - t))
    printf("-1\n");
  else {
    int j = 0;
    int l = n - t;
    while (l > 0 && j < po) {
      s3[pos[j]] = s1[pos[j]];
      j++;
      l--;
    }
    for (i = 0; i < n; i++) {
      if (l > 0) {
        while (s3[i] != '1') i++;
        s3[i] = s1[i];
        while (s3[i] != '1') i++;
        s3[i] = s2[i];
        l--;
        i--;
      } else if (s3[i] == '1') {
        char ch = 'a';
        while (ch == s1[i] || ch == s2[i]) ch++;
        s3[i] = ch;
      }
    }
    s3[n] = '\0';
    printf("%s\n", s3);
    return 0;
  }
}