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;
  int t;
  cin >> n;
  cin >> t;
  string a, b;
  cin >> a >> b;
  int p = 0;
  char c[a.length()];
  c[a.length()] = '\0';
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) p++;
  }
  if (n - t <= p) {
    int r = n - t;
    for (int i = 0; i < n; i++) {
      if (r && a[i] == b[i]) {
        c[i] = a[i];
        r--;
      } else {
        for (char t = 'a'; t <= 'z'; t++) {
          if (t != a[i] && t != b[i]) {
            c[i] = t;
            break;
          }
        }
      }
    }
    cout << c;
  } else {
    if (n - p < 2 * (n - t - p)) {
      cout << -1;
    } else {
      int r = 2 * (n - t - p);
      for (int i = 0; i < n; i++) {
        if (a[i] == b[i]) {
          c[i] = a[i];
        } else {
          if (r) {
            if (r > (n - t - p)) {
              c[i] = a[i];
            } else
              c[i] = b[i];
            r--;
          } else {
            for (char t = 'a'; t <= 'z'; t++) {
              if (t != a[i] && t != b[i]) {
                c[i] = t;
                break;
              }
            }
          }
        }
      }
      cout << c;
    }
  }
}