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;
int n;
char s1[111111];
char s2[111111];
char res[111111];
int t, k, eq;
char gd(char u, char v) {
  int i;
  for (i = 0; i <= 25; i++) {
    if ('a' + i != u && 'a' + i != v) {
      return 'a' + i;
    }
  }
}
int main(void) {
  cin >> n >> t;
  k = n - t;
  scanf("%s", &s1[1]);
  scanf("%s", &s2[1]);
  int i;
  for (i = 1; i <= n; i++) {
    if (s1[i] == s2[i]) {
      eq++;
    }
  }
  if (eq >= k) {
    int cur = 0;
    for (i = 1; i <= n; i++) {
      if (s1[i] != s2[i] || cur >= k) {
        res[i] = gd(s1[i], s2[i]);
      } else {
        cur++;
        res[i] = s1[i];
      }
    }
    for (i = 1; i <= n; i++) {
      printf("%c", res[i]);
    }
    return 0;
  } else {
    if (n - eq < 2 * (k - eq)) {
      cout << -1;
      return 0;
    }
    int cur1 = 0, cur2 = 0;
    for (i = 1; i <= n; i++) {
      if (s1[i] == s2[i]) {
        res[i] = s1[i];
      } else if (cur1 < k - eq) {
        cur1++;
        res[i] = s1[i];
      } else if (cur2 < k - eq) {
        cur2++;
        res[i] = s2[i];
      } else {
        res[i] = gd(s1[i], s2[i]);
      }
    }
    for (i = 1; i <= n; i++) {
      printf("%c", res[i]);
    }
  }
}