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;
const int N = 1e5 + 2;
char str[N], st[N], ans[N];
bool mark[26];
int main() {
  int n, t, f, l, mil, i, j;
  scanf("%d %d", &n, &t);
  scanf("%s %s", str, st);
  mil = f = n - t;
  for (i = 0; i < n; i++)
    if (str[i] == st[i]) mil--, ans[i] = '1';
  if (mil > 0 && mil * 2 > t + mil)
    printf("-1\n");
  else {
    i = 0;
    while (f > 0 && i < n) {
      if (ans[i] == '1') ans[i] = str[i], f--;
      i++;
    }
    f = l = mil;
    for (i = 0; i < n; i++) {
      if (ans[i] >= 'a' && ans[i] <= 'z') continue;
      if (f > 0)
        ans[i] = str[i], f--;
      else if (l > 0)
        ans[i] = st[i], l--;
      else {
        mark[0] = mark[1] = mark[2] = false;
        mark[str[i] - 'a'] = mark[st[i] - 'a'] = true;
        if (!mark[0])
          ans[i] = 'a';
        else if (!mark[1])
          ans[i] = 'b';
        else
          ans[i] = 'c';
      }
    }
    printf("%s\n", ans);
  }
  return 0;
}