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>
int n, t, cnt, chk;
char a[100000], b[100000], c[100000];
int main() {
  scanf("%d %d\n", &n, &t);
  scanf("%s %s", a, b);
  for (int i = 0; i < n; i++)
    if (a[i] != b[i]) cnt++;
  if (cnt <= t) {
    for (int i = 0; i < n; i++)
      if (a[i] != b[i])
        for (int j = 'a'; j <= 'z'; j++)
          if (a[i] != j && b[i] != j) c[i] = j;
    for (int i = 0; i < n; i++) {
      if (c[i] != 0) continue;
      if (a[i] == b[i] && t > cnt) {
        cnt++;
        for (int j = 'a'; j <= 'z'; j++)
          if (a[i] != j) c[i] = j;
      } else
        c[i] = a[i];
    }
    for (int i = 0; i < n; i++) printf("%c", c[i]);
  } else if (cnt > t * 2)
    printf("-1");
  else {
    int flag = 0;
    for (int i = 0; i < n; i++) {
      if (flag == 0 && a[i] != b[i]) {
        c[i] = a[i];
        flag = 1;
      } else {
        if (a[i] == b[i]) continue;
        c[i] = b[i];
        flag = 0;
        cnt--;
      }
      if (cnt == t) break;
    }
    for (int i = 0; i < n; i++) {
      if (c[i] != 0) continue;
      if (a[i] != b[i])
        for (int j = 'a'; j <= 'z'; j++)
          if (a[i] != j && b[i] != j) c[i] = j;
    }
    for (int i = 0; i < n; i++) {
      if (c[i] != 0) continue;
      if (a[i] == b[i] && t > cnt) {
        cnt++;
        for (int j = 'a'; j <= 'z'; j++)
          if (a[i] != j) c[i] = j;
      } else
        c[i] = a[i];
    }
    for (int i = 0; i < n; i++) printf("%c", c[i]);
  }
  return 0;
}