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 find(char a, char b) {
  for (char f = 'a';; f++)
    if (f != a && f != b) return f;
}
int main() {
  int n;
  int t;
  cin >> n >> t;
  char* s1 = new char[n];
  char* s2 = new char[n];
  char* s3 = new char[n];
  cin >> s1;
  cin >> s2;
  int diff = 0;
  int d1 = 0, d2 = 0;
  for (int i = 0; i < n; i++)
    if (s1[i] != s2[i]) diff++;
  int same = n - diff;
  int me = n - t;
  if (me <= same) {
    for (int i = 0; i < n; i++) {
      if (s1[i] == s2[i] && me > 0) {
        s3[i] = s1[i];
        me--;
      } else
        s3[i] = find(s1[i], s2[i]);
    }
  } else {
    if (same + (n - same) / 2 >= me) {
      int m1 = 0, m2 = 0;
      int mx = me - same;
      for (int i = 0; i < n; i++) {
        if (s1[i] == s2[i])
          s3[i] = s1[i];
        else {
          if (m1 < mx) {
            s3[i] = s1[i];
            m1++;
          } else if (m2 < mx) {
            s3[i] = s2[i];
            m2++;
          } else
            s3[i] = find(s1[i], s2[i]);
        }
      }
    } else {
      cout << -1;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) cout << s3[i];
  return 0;
}