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;
string s1, s2;
int f(string x, string y) {
  int n = x.length();
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (x[i] != y[i]) {
      ans++;
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, t;
  cin >> n >> t >> s1 >> s2;
  int x = f(s1, s2);
  if (x > t) {
    string ans = "";
    int l = 0, r = 0;
    for (int i = 0; i < n; i++) {
      if (s1[i] == s2[i]) {
        ans += s1[i];
      } else {
        if (l < (x - t)) {
          ans += s1[i];
          l++;
        } else if (r < (x - t)) {
          ans += s2[i];
          r++;
        } else {
          for (char j = 'a'; j <= 'z'; j++) {
            if (j != s1[i] && j != s2[i]) {
              ans += j;
              break;
            }
          }
        }
      }
    }
    if (f(ans, s1) == t && f(ans, s2) == t) {
      cout << ans << "\n";
    } else {
      cout << "-1\n";
    }
  } else {
    int common = n - t, f = 0;
    for (int i = 0; i < n; i++) {
      if (f < common && s1[i] == s2[i]) {
        cout << s1[i];
        f++;
      } else {
        for (char j = 'a'; j <= 'z'; j++) {
          if (j != s1[i] && j != s2[i]) {
            cout << j;
            break;
          }
        }
      }
    }
  }
}