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() {
  long long t, i, j, k, l, m, n, p, q;
  long long a, b, c, d;
  string s, s1, ans;
  while (cin >> n >> m) {
    string rem = "abc";
    cin >> s >> s1;
    ans = "";
    p = n - m;
    ans = s1;
    bool flag[n];
    for (i = 0; i < n; i++) flag[i] = 0;
    q = 0;
    for (i = 0; i < n; i++) {
      if (s[i] == s1[i] && q < p) {
        q++;
        flag[i] = 1;
      }
    }
    if (q == p) {
      for (i = 0; i < n; i++) {
        if (!flag[i]) {
          for (j = 0; j < 3; j++) {
            if (s[i] != rem[j] && s1[i] != rem[j]) {
              ans[i] = rem[j];
              break;
            }
          }
        }
      }
      cout << ans << endl;
    } else {
      a = n - q;
      b = p - q;
      if (a / 2 >= b) {
        j = 0;
        for (i = 0; j < b; i++) {
          if (!flag[i]) {
            flag[i] = 1;
            ans[i] = s[i];
            j++;
          }
        }
        j = 0;
        for (i = 0; j < b; i++) {
          if (!flag[i]) {
            flag[i] = 1;
            ans[i] = s1[i];
            j++;
          }
        }
        for (i = 0; i < n; i++) {
          if (!flag[i]) {
            for (j = 0; j < 3; j++) {
              if (s[i] != rem[j] && s1[i] != rem[j]) {
                ans[i] = rem[j];
                break;
              }
            }
          }
        }
        cout << ans << endl;
      } else
        cout << "-1\n";
    }
  }
  return 0;
}