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 s = "abcdefghijklmnopqrstuvwxyz";
char find1(char a, char b) {
  int i;
  for (i = 0; i < 26; i++) {
    if ((s[i] != a) && (s[i] != b)) return s[i];
  }
}
int main() {
  int n, t, i, j;
  string a, b;
  cin >> n >> t;
  cin >> a >> b;
  char ans[n + 3];
  int x = 0;
  for (i = 0; i < n; i++) {
    if (a[i] != b[i]) {
      ++x;
    }
  }
  if (t < (x + 1) / 2) {
    cout << "-1\n";
    return 0;
  } else if (t <= x) {
    int v = 2 * (x - t);
    for (i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        if (v > (x - t)) {
          ans[i] = a[i];
          --v;
        } else if (v > 0) {
          ans[i] = b[i];
          --v;
        } else {
          ans[i] = find1(a[i], b[i]);
        }
      } else {
        ans[i] = a[i];
      }
    }
  } else {
    int y = t - x;
    for (i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        ans[i] = find1(a[i], b[i]);
      }
    }
    for (i = 0; i < n; i++) {
      if (a[i] == b[i])
        if (y) {
          ans[i] = find1(a[i], b[i]);
          --y;
        } else {
          ans[i] = a[i];
        }
    }
  }
  ans[n] = '\0';
  cout << ans << endl;
  return 0;
}