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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    int n, t;
    cin >> n >> t;
    string s1, s2;
    cin >> s1 >> s2;
    char s3[n];
    for (int i = 0; i < n; i++) s3[i] = '0';
    vector<int> eq;
    for (int i = 0; i < n; i++)
      if (s1.at(i) == s2.at(i)) eq.push_back(i);
    int i = 0;
    int change = n - t;
    while (change && i < eq.size()) {
      s3[eq[i]] = s1.at(eq[i]);
      i++;
      change--;
    }
    if (change * 2 > n - eq.size()) {
      cout << -1;
      return 0;
    }
    i = 0;
    int x = change;
    while (change && i < n) {
      if (s3[i] == '0') {
        s3[i] = s1.at(i);
        i++;
        change--;
      } else {
        i++;
        continue;
      }
    }
    i = 0;
    while (x && i < n) {
      if (s3[i] == '0') {
        s3[i] = s2.at(i);
        i++;
        x--;
      } else {
        i++;
        continue;
      }
    }
    for (int i = 0; i < n; i++) {
      if (s3[i] == '0') {
        for (char j = 'a'; j <= 'z'; j++)
          if (j != s1.at(i) && j != s2.at(i)) {
            cout << j;
            break;
          }
      } else
        cout << s3[i];
    }
  }
}