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 main() {
  int n, t;
  scanf("%d %d", &n, &t);
  getchar();
  char s1[100001], s2[100001];
  gets(s1);
  gets(s2);
  int k = n - t, e = 0;
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) e++;
  }
  if (k <= e) {
    for (int i = 0; i < n; i++) {
      if (s1[i] == s2[i] && k > 0) {
        k--;
        putchar(s1[i]);
      } else {
        for (int j = 0; j < 26; j++) {
          if (('a' + j) != s1[i] && ('a' + j) != s2[i]) {
            putchar('a' + j);
            break;
          }
        }
      }
    }
  } else {
    if (n - e < 2 * (k - e))
      printf("-1");
    else {
      int a1 = k - e, a2 = k - e;
      for (int i = 0; i < n; i++) {
        if (s1[i] == s2[i])
          putchar(s1[i]);
        else {
          if (a1 > 0) {
            a1--;
            putchar(s1[i]);
          } else if (a2 > 0) {
            a2--;
            putchar(s2[i]);
          } else {
            for (int j = 0; j < 26; j++) {
              if (('a' + j) != s1[i] && ('a' + j) != s2[i]) {
                putchar('a' + j);
                break;
              }
            }
          }
        }
      }
    }
  }
  return 0;
}