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() {
  int n, t;
  cin >> n >> t;
  string s, z;
  cin >> s >> z;
  int dif = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] != z[i]) dif++;
  }
  if (dif > 2 * t) return cout << -1, 0;
  if (dif > t) {
    int noc = dif - t;
    noc *= 2;
    dif -= noc;
    int voc = noc;
    for (int i = 0; i < n; i++) {
      if (s[i] != z[i]) {
        if (dif) {
          if (s[i] != 'a' and z[i] != 'a')
            cout << 'a';
          else if (s[i] != 'b' and z[i] != 'b')
            cout << 'b';
          else if (s[i] != 'c' and z[i] != 'c')
            cout << 'c';
          dif--;
        } else {
          if (noc > voc / 2) {
            cout << s[i];
          } else {
            cout << z[i];
          }
          noc--;
        }
      } else
        cout << s[i];
    }
  } else {
    int noc = t - dif;
    dif;
    for (int i = 0; i < n; i++) {
      if (s[i] != z[i]) {
        if (dif) {
          if (s[i] != 'a' and z[i] != 'a')
            cout << 'a';
          else if (s[i] != 'b' and z[i] != 'b')
            cout << 'b';
          else if (s[i] != 'c' and z[i] != 'c')
            cout << 'c';
        }
      } else {
        if (noc) {
          if (s[i] == 'a')
            cout << "b";
          else
            cout << "a";
          noc--;
        } else
          cout << s[i];
      }
    }
  }
}