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;
const long long mx = 105;
const long long mod = 1e9 + 7;
const long long maxn = 1e17;
void solve() {
  long long n, i, j, k, t;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  bool ass = 0;
  long long tot = 0;
  string res;
  for (i = 0; i < n; i++) {
    res += ' ';
  }
  t = n - t;
  bool av[n];
  memset(av, 0, sizeof(av));
  for (i = 0; i < n; i++) {
    if (tot < t) {
      if (a[i] == b[i]) {
        res[i] = a[i];
        av[i] = true;
        tot++;
      }
    }
  }
  for (i = 0; i < n; i++) {
    if (tot < t) {
      if (a[i] != b[i]) {
        if (ass == 0) {
          res[i] = a[i];
          ass = !ass;
        } else {
          res[i] = b[i];
          tot++;
          ass = !ass;
        }
      }
    } else if (!av[i]) {
      if (a[i] == b[i]) {
        int v = (a[i] - 'a' + 1) % 26;
        res[i] = (v + 'a');
      } else {
        int v = (a[i] - 'a' + 1) % 26;
        if (v + 'a' != b[i]) {
          res[i] = (v + 'a');
        } else {
          v = (b[i] - 'a' + 1) % 26;
          res[i] = (v + 'a');
        }
      }
    }
  }
  if (tot < t) {
    cout << "-1\n";
  } else {
    cout << res << "\n";
  }
}
int main() {
  long long t = 1;
  while (t--) {
    solve();
  }
}