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, k, cnt = 0;
  cin >> n >> k;
  string s1, s2;
  cin >> s1;
  cin >> s2;
  vector<bool> eq(n, 0);
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      eq[i] = 1;
      cnt++;
    }
  }
  int diff = n - k;
  if (diff <= cnt) {
    for (int i = 0; i < n; i++) {
      if (diff && eq[i]) {
        cout << s1[i];
        diff--;
      } else {
        for (int j = 0; j < 26; j++) {
          char c = 'a' + j;
          if (c != s1[i] && c != s2[i]) {
            cout << c;
            break;
          }
        }
      }
    }
    return 0;
  }
  if (n - cnt < 2 * (diff - cnt)) {
    cout << -1 << endl;
    return 0;
  }
  int left1 = diff - cnt, left2 = diff - cnt;
  for (int i = 0; i < n; i++) {
    if (eq[i]) {
      cout << s1[i];
    } else if (left1) {
      left1--;
      cout << s1[i];
    } else if (left2) {
      left2--;
      cout << s2[i];
    } else {
      for (int j = 0; j < 26; j++) {
        char c = 'a' + j;
        if (c != s1[i] && c != s2[i]) {
          cout << c;
          break;
        }
      }
    }
  }
  cin.get();
  cin.get();
  return 0;
}