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() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, t;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  char ans[n + 5];
  for (int i = 0; i < n; i++) ans[i] = '0';
  int p = 0;
  vector<int> ind_diff, ind_equal;
  for (int i = 0; i < n; i++) {
    if (a[i] != b[i]) {
      p++;
      ind_diff.push_back(i);
    } else
      ind_equal.push_back(i);
  }
  if (t < (p + 1) / 2) {
    cout << -1;
    return 0;
  }
  int need = min(t, p);
  t -= need;
  need = p - need;
  for (int i = 0; i < need; i++) {
    ans[ind_diff[i]] = a[ind_diff[i]];
    ans[ind_diff[p - 1 - i]] = b[ind_diff[p - 1 - i]];
  }
  for (int i = need; i < p - need; i++) {
    for (int c = 'a'; c <= 'z'; c++) {
      if (c != a[ind_diff[i]] and c != b[ind_diff[i]]) {
        ans[ind_diff[i]] = c;
        break;
      }
    }
  }
  for (int i = 0; i < t; i++) {
    for (int c = 'a'; c <= 'z'; c++) {
      if (c != a[ind_equal[i]]) {
        ans[ind_equal[i]] = c;
        break;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (ans[i] == '0') ans[i] = a[i];
    cout << ans[i];
  }
  return 0;
}