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;
const int MOD = 1000000007;
long double PI = 4 * atan(1);
int n, t, cnt, req, rem, cur;
string a, b;
char ans[100001];
char calc(char x, char y) {
  char c;
  for (int i = 0; i < 26; i++) {
    c = ('a' + i);
    if (c != x && c != y) return c;
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> t >> a >> b;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) cnt++;
  }
  req = n - t;
  if (req <= cnt) {
    for (int i = 0; i < n; i++) {
      if (a[i] == b[i]) {
        if (req <= 0) {
          ans[i] = calc(a[i], b[i]);
        } else {
          ans[i] = a[i];
          req--;
        }
      } else {
        ans[i] = calc(a[i], b[i]);
      }
    }
  } else {
    rem = req - cnt;
    if ((n - cnt) < 2 * rem) {
      cout << -1;
      return 0;
    }
    for (int i = 0; i < n; i++) {
      if (a[i] == b[i])
        ans[i] = a[i];
      else {
        cur++;
        if (cur > rem && cur <= (2 * rem)) {
          ans[i] = b[i];
        } else if (cur <= rem) {
          ans[i] = a[i];
        } else {
          ans[i] = calc(a[i], b[i]);
        }
      }
    }
  }
  for (int i = 0; i < n; i++) cout << ans[i];
  return 0;
}