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>
char three(char a, char b) {
  for (int i = 0; i < 26; i++) {
    if (i + 'a' != a && i + 'a' != b) return i + 'a';
  }
}
int main() {
  bool o = 0;
  int n, t;
  char s[3][100005];
  scanf("%d%d", &n, &t);
  scanf("%s", s[0]);
  scanf("%s", s[1]);
  int p = 0, q = 0;
  for (int i = 0; i < n; i++) {
    if (s[0][i] != s[1][i])
      p++;
    else
      q++;
  }
  int r = p % 2;
  p -= r;
  if (t < p / 2 + r)
    printf("-1\n");
  else {
    if (t <= p + r) {
      int d = (t - p / 2 - r) * 2;
      p -= d;
      d += r;
      for (int i = 0; i < n; i++) {
        if (s[0][i] != s[1][i]) {
          if (d) {
            s[2][i] = three(s[0][i], s[1][i]);
            d--;
          } else {
            s[2][i] = s[p % 2][i];
            p--;
          }
        } else {
          s[2][i] = s[0][i];
        }
      }
    } else {
      int d = t - p - r;
      for (int i = 0; i < n; i++) {
        if (s[0][i] != s[1][i]) {
          s[2][i] = three(s[0][i], s[1][i]);
        } else if (d) {
          s[2][i] = three(s[0][i], s[1][i]);
          d--;
        } else {
          s[2][i] = s[0][i];
        }
      }
    }
    printf("%s\n", s[2]);
  }
  return 0;
}