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, t, cnt = 0;
  string s1, s2;
  cin >> n >> t;
  int H = t;
  cin >> s1 >> s2;
  for (int i = 0; i < n; ++i)
    if (s1[i] != s2[i]) cnt++;
  int T = t;
  string ans = s1;
  int k = 0;
  for (int i = 0; i < n; ++i) {
    if (s1[i] != s2[i]) {
      if (cnt <= T) {
        if (s1[i] != 'a' && s2[i] != 'a')
          ans[i] = 'a';
        else if (s1[i] != 'b' && s2[i] != 'b')
          ans[i] = 'b';
        else
          ans[i] = 'c';
        T--;
        cnt--;
      } else {
        if (k % 2 == 0)
          ans[i] = s1[i];
        else
          ans[i] = s2[i];
        k++;
        if (k == 2) {
          k = 0;
          T--;
          cnt -= 2;
        }
      }
    }
  }
  for (int i = 0; i < n && T > 0; ++i)
    if (s1[i] == s2[i] && ans[i] == s1[i]) {
      if (s1[i] == 'a')
        ans[i] = 'b';
      else
        ans[i] = 'a';
      T--;
    }
  int k1 = 0, k2 = 0;
  for (int i = 0; i < n; ++i)
    if (s1[i] != ans[i]) k1++;
  for (int i = 0; i < n; ++i)
    if (s2[i] != ans[i]) k2++;
  if (k1 != H || k2 != H) {
    cout << -1 << endl;
    return 0;
  }
  cout << ans << endl;
}