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;
map<char, int> ma;
map<int, bool> maa;
int main() {
  int n, m;
  cin >> n >> m;
  string a, b;
  cin >> a >> b;
  string res(n, 0);
  int c1 = a.size() - m;
  int c2 = b.size() - m;
  for (int i = 0; i < min(a.size(), b.size()) && i < n; i++) {
    if (a[i] == b[i]) {
      if (c1 && c2) {
        res[i] = a[i];
        c1--;
        c2--;
        maa[i]++;
      } else {
        break;
      }
    }
  }
  if (b.size() < a.size()) {
    swap(a, b);
  }
  int it = 0;
  while (c1 && it < n) {
    if (!maa[it] && res[it] == 0) {
      res[it] = a[it];
      c1--;
    }
    it++;
  }
  it = 0;
  while (c2 && it < n) {
    if (!maa[it] && res[it] == 0) {
      res[it] = b[it];
      c2--;
    }
    it++;
  }
  if (c1 > 0 || c2 > 0) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (res[i] == 0)
      for (int j = 'a'; j <= 'z'; j++) {
        if (i < b.size()) {
          if (i < a.size()) {
            if (j != a[i] && j != b[i]) {
              res[i] = j;
              break;
            }
          } else {
            if (j != b[i]) {
              res[i] = j;
              break;
            }
          }
        } else if (i < a.size()) {
          if (i < b.size()) {
            if (j != a[i] && j != b[i]) {
              res[i] = j;
              break;
            }
          } else {
            if (j != a[i]) {
              res[i] = j;
              break;
            }
          }
        }
      }
  }
  cout << res << endl;
}