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;
int main() {
  int n, m, i, j;
  cin >> n >> m;
  string s, t;
  cin >> s >> t;
  int w = m;
  string d = "";
  for (i = 0; i < n; i++) {
    if (s[i] == t[i]) {
      d += s[i];
    } else {
      if (w > 0) {
        d += s[i];
      } else {
        d += t[i];
      }
      w--;
      if (-w > m) {
        puts("-1");
        return 0;
      }
    }
  }
  int c = m;
  for (i = 0; i < n; i++) {
    if (d[i] != t[i]) c--;
  }
  if (c > 0) {
    for (i = 0; i < n; i++) {
      if (s[i] != t[i]) continue;
      for (char k = 'a'; k <= 'z'; k++) {
        if (s[i] != k && t[i] != k) {
          d[i] = k;
          break;
        }
      }
      c--;
      if (c == 0) break;
    }
  }
  if (c > 0) {
    puts("-1");
    return 0;
  }
  c = m;
  for (i = 0; i < n; i++) {
    if (d[i] != s[i]) c--;
  }
  if (c > 0) {
    for (i = 0; i < n; i++) {
      if (s[i] == d[i] && d[i] != t[i]) {
        for (char k = 'a'; k <= 'z'; k++) {
          if (s[i] != k && t[i] != k) {
            d[i] = k;
            break;
          }
        }
        c--;
        if (c == 0) break;
      }
    }
  }
  if (c > 0)
    puts("-1");
  else
    cout << d;
}