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;
char diffchar(char c1, char c2) {
  char c;
  if (c2 < c1) swap(c1, c2);
  if (abs(c1 - c2) <= 1) {
    c = 'z';
    if (c == c2) c = 'a';
  } else {
    c = (c1 + c2) / 2;
  }
  return c;
}
int main() {
  long long n, t;
  cin >> n >> t;
  string s1, s2;
  cin >> s1 >> s2;
  long long diff = 0;
  long long same = 0;
  long long i;
  string ans(n, '@');
  for (i = 0; i < n; i++) {
    if (s1[i] == s2[i]) same++;
    if (s1[i] != s2[i]) diff++;
  }
  if (same >= n - t) {
    same = 0;
    for (i = 0; i < n; i++) {
      if (same < n - t && s1[i] == s2[i]) {
        same++;
        ans[i] = s1[i];
      } else {
        ans[i] = diffchar(s1[i], s2[i]);
      }
    }
  } else {
    long long remain1 = n - t - same;
    long long remain2 = remain1;
    for (i = 0; i < n; i++) {
      if (s1[i] == s2[i]) {
        ans[i] = s1[i];
      }
    }
    if (2 * remain1 > n - same) {
      ans = "-1";
    } else {
      for (i = 0; i < n; i++) {
        if (ans[i] == '@') {
          if (remain1) {
            ans[i] = s1[i];
            remain1--;
          } else if (remain2) {
            ans[i] = s2[i];
            remain2--;
          } else {
            ans[i] = diffchar(s1[i], s2[i]);
          }
        }
      }
    }
  }
  cout << ans;
  return 0;
}