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>
int n, t, m, x, m1, m2;
char a[111111], b[111111], c[111111];
int same[111111];
char diff(char x, char y) {
  for (char i = 'a'; i <= 'z'; i++)
    if (x != i && y != i) return i;
  return '?';
}
int main() {
  scanf("%d%d", &n, &t);
  scanf("%s%s", a, b);
  m = n - t;
  x = 0;
  for (int i = 0; i < n; i++)
    if (a[i] == b[i]) same[x++] = i;
  if (x >= m) {
    for (int i = 0; i < m; i++) c[same[i]] = a[same[i]];
    for (int i = 0; i < n; i++)
      if (!c[i]) c[i] = diff(a[i], b[i]);
  } else if ((n - x) < (m - x) * 2) {
    printf("-1\n");
    return 0;
  } else {
    for (int i = 0; i < x; i++) c[same[i]] = a[same[i]];
    m1 = m2 = m - x;
    for (int i = 0; i < n; i++) {
      if (c[i]) continue;
      if (m1) {
        c[i] = a[i];
        m1--;
      } else if (m2) {
        c[i] = b[i];
        m2--;
      } else {
        c[i] = diff(a[i], b[i]);
      }
    }
  }
  printf("%s\n", c);
  return 0;
}