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 long long INF = 1e12;
const int MAXN = 1e5 + 1, MODM = 1e9 + 7, P = 1e6 + 3, MULT1 = 30, MULT2 = 239,
          LUCKY = 13, ELEVEN = 11, MAXM = 101;
int main() {
  int n, t;
  cin >> n >> t;
  string s, s1;
  cin >> s >> s1;
  int k = 0;
  vector<int> ans;
  for (int i = 0; i < n; i++)
    if (s[i] == s1[i]) {
      k++;
      ans.push_back(i);
    }
  if (t < (n - k) / 2) {
    cout << -1;
  } else {
    vector<char> mas(n);
    if (k >= n - t) {
      k = n - t;
      for (int i = 0; i < n; i++)
        if (s[i] == s1[i] && k > 0) {
          k--;
          mas[i] = s[i];
        } else {
          for (char c = 'a'; c <= 'z'; c++)
            if (s[i] != c && s1[i] != c) {
              mas[i] = c;
              break;
            }
        }
      for (int i = 0; i < n; i++) cout << mas[i];
    } else {
      for (int i = 0; i < n; i++)
        if (s[i] == s1[i]) mas[i] = s[i];
      int l = n - t - k;
      if (n - k < 2 * l) {
        cout << -1;
        return 0;
      } else {
        int j = 0;
        for (int i = 0; i < n; i++)
          if (s[i] != s1[i]) {
            if (j % 2 == 0 && l > 0) {
              mas[i] = s[i];
              j++;
            } else if (j % 2 == 1 && l > 0) {
              mas[i] = s1[i];
              j++;
              l--;
            } else {
              for (char c = 'a'; c <= 'z'; c++)
                if (s[i] != c && s1[i] != c) {
                  mas[i] = c;
                  break;
                }
            }
          }
      }
      for (int i = 0; i < n; i++) cout << mas[i];
    }
  }
  return 0;
}