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, i, d, d1, d2, o, c;
  string a, b;
  char ans[100005];
  cin >> n >> t;
  cin >> a >> b;
  d = 0;
  for (i = 0; i < n; i++)
    if (a[i] != b[i]) d++;
  if (t < (d + 1) / 2) {
    cout << "-1\n";
    return 0;
  }
  o = 2 * t - d;
  if (o > d) {
    c = t - d;
    o = d;
    d1 = d2 = 0;
  } else {
    c = 0;
    d1 = d2 = t - o;
  }
  for (i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      if (c) {
        ans[i] = a[i] == 'a' ? 'b' : 'a';
        c--;
      } else {
        ans[i] = a[i];
      }
    } else {
      if (d1) {
        d1--;
        ans[i] = b[i];
      } else if (d2) {
        d2--;
        ans[i] = a[i];
      } else if (o) {
        o--;
        if (a[i] != 'a' && b[i] != 'a')
          ans[i] = 'a';
        else if (a[i] != 'b' && b[i] != 'b')
          ans[i] = 'b';
        else
          ans[i] = 'c';
      }
    }
  }
  ans[n] = '\0';
  cout << ans << endl;
  return 0;
}