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;
int32_t main() {
  long long n, k;
  string s1, s2, s3;
  long long c1 = 0, c2 = 0, diff = 0;
  cin >> n >> k;
  cin >> s1 >> s2;
  for (long long i = 0; i < n; i++)
    if (s1[i] != s2[i]) diff++;
  if ((diff + 1) / 2 > k) {
    cout << "-1\n";
    return 0;
  }
  diff = 0, s3 = s1;
  for (long long i = 0; i < n; i++) {
    for (char j = 'a'; j <= 'd'; j++) {
      if (s1[i] != s2[i] && s1[i] != j && s2[i] != j) {
        s3[i] = j;
        diff++;
        c1++;
        c2++;
        break;
      }
    }
  }
  for (long long i = 0; i < n && diff < k; i++) {
    for (char j = 'a'; j <= 'd'; j++) {
      if (s1[i] == s2[i] && s1[i] != j) {
        s3[i] = j;
        c1++;
        c2++;
        diff++;
        break;
      }
    }
  }
  for (long long i = 0; i < n && (c1 > k || c2 > k); i++) {
    for (char j = 'a'; j <= 'd'; j++) {
      if (s1[i] != s2[i] && s2[i] != j) {
        if (c1 > c2) {
          s3[i] = s1[i];
          c1--;
        } else {
          s3[i] = s2[i];
          c2--;
        }
        break;
      }
    }
  }
  if (c1 == k && c2 == k)
    cout << s3;
  else
    cout << "-1";
}