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, t;
char s1[100010], s2[100010];
char ans[100010];
int main() {
  int i, tmp, tmp_a, tmp_b;
  char y;
  tmp = 0;
  scanf("%d%d", &n, &t);
  scanf("%s", s1);
  scanf("%s", s2);
  t = n - t;
  for (i = 0; i < n; i++)
    if (s1[i] == s2[i]) {
      ans[i] = s1[i];
      tmp++;
    }
  tmp_a = tmp_b = tmp;
  if (tmp >= t) {
    for (i = 0; i < n; i++) {
      if (s1[i] != s2[i]) {
        for (y = 'a'; y <= 'z'; y++) {
          if (y != s1[i] && y != s2[i]) {
            ans[i] = y;
            break;
          }
        }
      }
      if (s1[i] == s2[i] && tmp > t) {
        for (y = 'a'; y <= 'z'; y++) {
          if (y != s1[i] && y != s2[i]) {
            ans[i] = y;
            break;
          }
        }
        tmp--;
      }
    }
    ans[n] = '\0';
    printf("%s\n", ans);
  } else {
    for (i = 0; i < n; i++) {
      if (s1[i] == s2[i])
        continue;
      else {
        if (tmp_a < t) {
          tmp_a++;
          ans[i] = s1[i];
        } else if (tmp_b < t) {
          tmp_b++;
          ans[i] = s2[i];
        } else {
          for (y = 'a'; y <= 'z'; y++) {
            if (y != s1[i] && y != s2[i]) {
              ans[i] = y;
              break;
            }
          }
        }
      }
    }
    if (tmp_a == tmp_b && tmp_a == t) {
      ans[n] = '\0';
      printf("%s\n", ans);
    } else
      printf("-1\n");
  }
  return 0;
}