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>
void change(char a[], char b[], int i) {
  a[i] += 1;
  if (a[i] == b[i]) a[i] += 1;
  if (a[i] > 'z') {
    a[i] = 'a';
    if (a[i] == b[i]) a[i] += 1;
  }
}
void convert(char a[], char b[], int n, int t, int s) {
  int i, c, d, e[100001], k, l;
  d = n - s;
  c = 0;
  l = 0;
  for (i = 0; i < n; i++) {
    if (a[i] != b[i]) {
      if (d <= t) {
        change(a, b, i);
        d--;
        t--;
      } else if (c == 0)
        c = 1;
      else if (c == 1) {
        a[i] = b[i];
        e[l++] = i;
        t--;
        d -= 2;
        c = 0;
      }
    }
  }
  k = 0;
  for (i = 0; i < n; i++)
    if (((k < l) && i != e[k]) || (k >= l)) {
      if (a[i] == b[i]) {
        if (t > 0) {
          change(a, b, i);
          t--;
        } else
          break;
      }
    } else
      k++;
}
int main() {
  char a[100001], b[100001];
  int n, t, s = 0, i;
  scanf("%d%d", &n, &t);
  scanf("%*c%s", a);
  scanf("%*c%s", b);
  for (i = 0; i < n; i++)
    if (a[i] == b[i]) s++;
  if (t > n / 2) {
    convert(a, b, n, t, s);
    printf("%s\n", a);
  } else {
    if (s >= (n - 2 * t)) {
      convert(a, b, n, t, s);
      printf("%s\n", a);
    } else
      printf("-1\n");
  }
  return 0;
}