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>
long long int mod = 1000000007;
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n, t;
  cin >> n >> t;
  string s1, s2;
  cin >> s1;
  cin >> s2;
  int a = 0, b = 0;
  int x = (n - t);
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      a++;
    } else {
      b++;
    }
  }
  string ans = "";
  if (x <= a) {
    for (int i = 0; i < n; i++) {
      if (s1[i] == s2[i] && x > 0) {
        ans += s1[i];
        x--;
      } else {
        for (char c = 'a'; c <= 'z'; c++) {
          if (c != s1[i] && c != s2[i]) {
            ans += c;
            break;
          }
        }
      }
    }
    cout << ans << endl;
    return 0;
  }
  if (2 * (x - a) > (n - a)) {
    cout << -1 << endl;
    return 0;
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      ans += s1[i];
    } else {
      if (cnt < (x - a)) {
        ans += s1[i];
        cnt++;
      } else if (cnt < (2 * (x - a))) {
        ans += s2[i];
        cnt++;
      } else {
        for (char c = 'a'; c <= 'z'; c++) {
          if (c != s1[i] && c != s2[i]) {
            ans += c;
            break;
          }
        }
      }
    }
  }
  cout << ans << endl;
  return 0;
}