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, t;
  cin >> n >> t;
  string a, b, ans;
  cin >> a >> b;
  ans = a;
  int same(0), diff(0);
  for (long p = 0; p < n; p++) {
    if (a[p] == b[p]) {
      ++same;
    } else {
      ++diff;
    }
  };
  if (2 * t < diff) {
    cout << -1;
    return 0;
  }
  int changeSame(0), changeDiff(0), changeFromA(0);
  if (t > diff) {
    changeSame = t - diff;
    changeDiff = diff;
  } else {
    changeDiff = 2 * t - diff;
    changeFromA = t - changeDiff;
  }
  for (int p = 0; p < n; p++) {
    if (a[p] == b[p]) {
      if (changeSame) {
        --changeSame;
        ans[p] = (a[p] == 'a') ? 'b' : 'a';
      } else
        ans[p] = a[p];
    } else if (changeDiff) {
      --changeDiff;
      ans[p] = 'a';
      while (ans[p] == a[p] || ans[p] == b[p]) {
        ++ans[p];
      }
    } else if (changeFromA) {
      --changeFromA;
      ans[p] = b[p];
    } else {
      ans[p] = a[p];
    }
  }
  cout << ans << endl;
}