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>
double eps = 1e-9;
using namespace std;
char chk(char a, char b, int t) {
  if (t == 3) {
    for (int i = 0; i < 26; i++) {
      if (char(i + 97) != a && char(i + 97) != b) {
        return char(i + 97);
      }
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  int n, t;
  cin >> n >> t;
  string a, b;
  cin >> a >> b;
  string c;
  int diff = 0, same = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      same++;
    } else {
      diff++;
    }
  }
  for (int i = 0; i <= diff; i++) {
    if ((diff - i) % 2 == 0) {
      int k = t - ((diff + i) / 2);
      if (k > same || k < 0) {
        continue;
      }
      int s = i;
      int g = (diff - i) / 2;
      int h = (diff - i) / 2;
      for (int j = 0; j < n; j++) {
        if (a[j] != b[j]) {
          if (s > 0) {
            c.push_back(chk(a[j], b[j], 3));
            s--;
          } else {
            if (g > 0) {
              c.push_back(a[j]);
              g--;
            } else {
              c.push_back((b[j]));
            }
          }
        } else {
          if (k > 0) {
            c.push_back(chk(a[j], b[j], 3));
            k--;
          } else {
            c.push_back(a[j]);
          }
        }
      }
      cout << c;
      return 0;
    }
  }
  cout << -1;
  return 0;
}