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 str1[100009];
char str2[100009];
char ans[100009];
int main() {
  int n, t, x = 0, y = 0;
  scanf("%d%d", &n, &t);
  scanf("%s", str1);
  scanf("%s", str2);
  for (int i = 0; i < n; i++) {
    if (str1[i] != str2[i]) x++;
  }
  int rem = t - x;
  if (t >= x) {
    for (int i = 0; i < n; i++) {
      if (str1[i] != str2[i]) {
        int q = ((str1[i] - 'a') + 1) % 26;
        if (q == (str2[i] - 'a')) {
          q++;
          q %= 26;
        }
        char c = 'a' + q;
        ans[i] = c;
      }
    }
    for (int i = 0; i < n; i++) {
      if (str1[i] == str2[i]) {
        if (rem <= 0)
          ans[i] = str1[i];
        else {
          int q = ((str1[i] - 'a') + 1) % 26;
          char c = 'a' + q;
          ans[i] = c;
          rem--;
        }
      }
    }
  } else {
    if (2 * t < x) {
      printf("-1\n");
      return 0;
    } else {
      int p = x - t;
      int flag = 0;
      for (int i = 0; i < n; i++) {
        if (str1[i] != str2[i]) {
          if (flag == 2) {
            int q = ((str1[i] - 'a') + 1) % 26;
            if (q == (str2[i] - 'a')) {
              q++;
              q %= 26;
            }
            char c = 'a' + q;
            ans[i] = c;
            continue;
          }
          p--;
          if (flag == 0) ans[i] = str1[i];
          if (flag == 1) ans[i] = str2[i];
          if (p == 0) {
            p = x - t;
            flag++;
          }
        } else
          ans[i] = str1[i];
      }
    }
  }
  for (int i = 0; i < n; i++) printf("%c", ans[i]);
  printf("\n");
}