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;
char getdiff(char a, char b) {
  char x = a;
  while (x == a || x == b) {
    x = (((x - 'a') + 1) % 26) + 'a';
  }
  return x;
}
void solve() {
  int n, k;
  cin >> n >> k;
  string s, t;
  cin >> s >> t;
  int diff = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] != t[i]) diff++;
  }
  int mindif = diff / 2 + (diff % 2);
  if (k < mindif) {
    cout << -1 << endl;
    return;
  }
  vector<char> ans(n, 'Z');
  int tochange = min(diff, k);
  k -= tochange;
  int keepsame = diff - tochange;
  for (int i = 0, j = 0; i < n; i++) {
    if (j >= keepsame) break;
    if (s[i] != t[i]) {
      ans[i] = s[i];
      j++;
    }
  }
  for (int i = n - 1, j = 0; i >= 0; i--) {
    if (j >= keepsame) break;
    if (s[i] != t[i]) {
      ans[i] = t[i];
      j++;
    }
  }
  for (int i = 0; i < n; i++) {
    if (s[i] != t[i] && ans[i] == 'Z') ans[i] = getdiff(s[i], t[i]);
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == t[i]) {
      if (k > 0) {
        ans[i] = getdiff(s[i], t[i]);
      } else {
        ans[i] = s[i];
      }
      k--;
    }
  }
  for (auto x : ans) cout << x;
  cout << endl;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  solve();
}