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;
char a[100005], b[100005], ans[100005];
char C[30] = "abcdefghijklmnopqrstuvwxyz";
int main() {
  int n, t;
  scanf("%d%d", &n, &t);
  scanf("%s %s", a, b);
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] != b[i]) cnt++;
  }
  if (cnt % 2 == 1 && t <= cnt / 2) {
    printf("-1\n");
  } else if (cnt % 2 == 0 && t < cnt / 2)
    printf("-1\n");
  else {
    if (t >= cnt) {
      for (int i = 0; i < n; i++) {
        if (a[i] != b[i]) {
          int j = 0;
          while (C[j] == a[i] || C[j] == b[i]) j++;
          ans[i] = C[j];
          t--;
        } else
          ans[i] = a[i];
      }
      for (int i = 0; i < n && t; i++) {
        if (a[i] == b[i]) {
          int j = 0;
          while (C[j] == a[i] || C[j] == b[i]) j++;
          ans[i] = C[j];
          t--;
        }
      }
    } else {
      strcpy(ans, b);
      int tmp = cnt - t;
      for (int i = 0; i < n && tmp; i++) {
        if (a[i] != b[i]) ans[i] = a[i], tmp--;
      }
      tmp = 2 * t - cnt;
      for (int i = 0; i < n && tmp; i++) {
        if (ans[i] != a[i] && ans[i] == b[i]) {
          int j = 0;
          while (C[j] == a[i] || C[j] == b[i]) j++;
          ans[i] = C[j];
          tmp--;
        }
      }
    }
    printf("%s\n", ans);
  }
  return 0;
}