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, flag[100010];
char s1[100010], s2[100010], s[100010];
int main() {
  srand(time(0));
  scanf("%d%d", &n, &t);
  scanf("%s%s", s1 + 1, s2 + 1);
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (s1[i] == s2[i]) {
      flag[i] = 1;
      cnt++;
    }
  if (n - t > (n + cnt) / 2) {
    printf("-1\n");
    return 0;
  }
  int curr = n - t;
  for (int i = 1; i <= n; i++)
    if (curr && flag[i]) {
      s[i] = s1[i];
      curr--;
    } else
      while (1) {
        int cur = rand() % 26 + 'a';
        if (s1[i] == cur || s2[i] == cur) continue;
        s[i] = cur;
        break;
      }
  if (!curr) {
    printf("%s", s + 1);
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (curr && !flag[i]) {
      int d = i + 1;
      while (flag[d]) d++;
      s[i] = s1[i];
      s[d] = s2[d];
      flag[d] = 1;
      curr--;
    }
  printf("%s", s + 1);
  return 0;
}