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, ti, i, k, diff = 0, ind, j;
  string s, t;
  cin >> n >> ti;
  cin >> s;
  cin >> t;
  string ans = s;
  for (i = 0; i < n; i++) {
    if (s[i] != t[i]) {
      diff++;
      ind = i;
    }
  }
  if (diff > 2 * ti) {
    cout << "-1" << endl;
    return 0;
  }
  if (diff <= ti) {
    for (i = 0; i < n; i++) {
      if (s[i] != t[i]) {
        for (j = 0; j < 26; j++) {
          if (s[i] != (j + 97) && t[i] != (j + 97)) {
            ans[i] = (char)(j + 97);
            ti--;
            break;
          }
        }
      }
    }
    if (ti) {
      for (i = 0; i < n; i++) {
        if (s[i] == t[i]) {
          k = s[i] - 97;
          k = (k + 1) % 26;
          ans[i] = (char)(k + 97);
          ti--;
        }
        if (ti == 0) break;
      }
    }
  } else {
    int c = diff - ti, c2;
    c2 = c;
    ti -= c;
    for (i = 0; i < n; i++) {
      if (s[i] != t[i]) {
        if (c != 0) {
          ans[i] = s[i];
          c--;
        } else {
          if (c2 != 0) {
            ans[i] = t[i];
            c2--;
          } else {
            for (j = 0; j < 26; j++) {
              if (s[i] != (j + 97) && t[i] != (j + 97)) {
                ans[i] = (char)(j + 97);
                ti--;
                break;
              }
            }
          }
        }
      }
    }
    if (ti) {
      for (i = 0; i < n; i++) {
        if (s[i] == t[i]) {
          k = s[i] - 97;
          k = (k + 1) % 26;
          ans[i] = (char)(k + 97);
          ti--;
        }
        if (ti == 0) break;
      }
    }
  }
  cout << ans << endl;
  return 0;
}