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;
vector<int> v;
int main() {
  int n, t;
  string s1, s2, ans;
  cin >> n >> t;
  cin >> s1 >> s2;
  v.resize(n);
  ans.resize(n);
  int c = 0;
  int kp = 0, kk = 0;
  t = n - t;
  for (int i = 0; i < n; i++)
    if (s1[i] == s2[i] && c < t) {
      c++;
      ans[i] = s1[i];
      v[i] = 1;
    } else
      v[i] = -1;
  t -= c;
  for (int i = 0; i < n; i++) {
    if (v[i] == 1) {
      continue;
    } else {
      if (s1[i] != s2[i] && kp < t) {
        kp++;
        ans[i] = s1[i];
        continue;
      }
      if (s1[i] != s2[i] && kk < t) {
        kk++;
        ans[i] = s2[i];
        continue;
      }
      if (s1[i] != 'a' && s2[i] != 'a') ans[i] = 'a';
      if (s1[i] != 'b' && s2[i] != 'b') ans[i] = 'b';
      if (s1[i] != 'z' && s2[i] != 'z') ans[i] = 'z';
    }
  }
  if (kk < t || kp < t) {
    cout << -1;
    return 0;
  }
  cout << ans;
  return 0;
}