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 s1[100010], s2[100010];
char ans[100010];
char fun(int c1, int c2) {
  for (int i = 0; i < 26; i++)
    if (i != c1 && i != c2) return i + 'a';
}
int main() {
  int n, t;
  scanf("%d %d", &n, &t);
  scanf("%s", s1);
  scanf("%s", s2);
  int x = 0;
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i]) x++;
  }
  if (x > 2 * t)
    printf("-1\n");
  else if (x <= t) {
    int tmpt = 0;
    for (int i = 0; i < n; i++) {
      if (s1[i] != s2[i]) {
        tmpt++;
        ans[i] = fun(s1[i] - 'a', s2[i] - 'a');
      }
    }
    for (int i = 0; i < n; i++) {
      if (s1[i] == s2[i]) {
        if (tmpt != t) {
          tmpt++;
          ans[i] = fun(s1[i] - 'a', s2[i] - 'a');
        } else
          ans[i] = s1[i];
      }
    }
    ans[n] = '\0';
    printf("%s\n", ans);
  } else {
    int tmpt = 0;
    int t1 = 2 * t - x, t2 = 2 * t - x;
    for (int i = 0; i < n; i++) {
      if (s1[i] == s2[i])
        ans[i] = s1[i];
      else {
        if (tmpt < 2 * t - x) {
          tmpt++;
          ans[i] = fun(s1[i] - 'a', s2[i] - 'a');
        } else if (t1 < t) {
          t1++;
          ans[i] = s2[i];
        } else if (t2 < t) {
          t2++;
          ans[i] = s1[i];
        }
      }
    }
    ans[n] = '\0';
    printf("%s\n", ans);
  }
  return 0;
}