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 a[100010], b[100010];
int c[100010];
char dif(int i) {
  char d = 'a';
  for (; d <= 'z'; d++) {
    if (d != a[i] && d != b[i]) return d;
  }
  return d;
}
int main() {
  int n, t, i;
  scanf("%d%d", &n, &t);
  scanf(" %s %s", a, b);
  int d = 0;
  memset(c, 0, sizeof(c));
  for (i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      d++;
      c[i] = 1;
    }
  }
  if (n - t > d + (n - d) / 2)
    printf("-1\n");
  else {
    int j = n - t;
    if (j > d) {
      int temp = j - d;
      for (i = 0; i < n && temp; i++) {
        if (!c[i]) {
          temp--;
          c[i] = 2;
        }
      }
      temp = j - d;
      for (i = 0; i < n && temp; i++) {
        if (!c[i]) {
          temp--;
          c[i] = 3;
        }
      }
    } else {
      for (i = 0; i < n && j; i++) {
        if (c[i]) j--;
      }
      for (; i < n; i++) {
        c[i] = 0;
      }
    }
    for (i = 0; i < n; i++) {
      if (c[i] == 2 || c[i] == 1)
        printf("%c", a[i]);
      else if (c[i] == 3)
        printf("%c", b[i]);
      else
        printf("%c", dif(i));
    }
  }
  return 0;
}