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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, t;
  cin >> n >> t;
  string s1, s2;
  cin >> s1 >> s2;
  int c = 0;
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      c++;
    }
  }
  int sum = c / 2 + (n - c);
  if (sum < n - t) {
    cout << -1 << endl;
    return 0;
  }
  char ans[100005];
  memset(ans, '0', sizeof ans);
  int cc = 0;
  for (int i = 0; i < n; i++) {
    if (cc >= n - t) {
      break;
    }
    if (s1[i] == s2[i]) {
      ans[i] = s1[i];
      cc++;
    }
  }
  cc = 0;
  for (int i = 0; i < n; i++) {
    if (cc >= (n - t) - (n - c)) {
      break;
    }
    if (s1[i] != s2[i]) {
      ans[i] = s1[i];
      cc++;
    }
  }
  cc = 0;
  for (int i = 0; i < n; i++) {
    if (cc >= (n - t) - (n - c)) {
      break;
    }
    if (s1[i] != s2[i] && ans[i] == '0') {
      ans[i] = s2[i];
      cc++;
    }
  }
  for (int i = 0; i < n; i++) {
    if (ans[i] == '0') {
      if (s1[i] != 'a' && s2[i] != 'a') {
        ans[i] = 'a';
      } else if (s1[i] != 'b' && s2[i] != 'b') {
        ans[i] = 'b';
      } else if (s1[i] != 'c' && s2[i] != 'c') {
        ans[i] = 'c';
      }
    }
  }
  ans[n] = '\0';
  cout << ans << endl;
  return 0;
}