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;
char s1[100001], s2[100001], s3[100001];
int n, t;
void complete_difer() {
  int i = 0;
  while (i < n) {
    if (s3[i] == 0) {
      if (s1[i] != s2[i]) {
        if (s1[i] == 'a')
          if (s2[i] == 'b')
            s3[i] = 'c';
          else
            s3[i] = 'b';
        else if (s2[i] == 'a')
          if (s1[i] == 'b')
            s3[i] = 'c';
          else
            s3[i] = 'b';
        else
          s3[i] = 'a';
      } else {
        if (s1[i] == 'a')
          s3[i] = 'z';
        else
          s3[i] = 'a';
      }
    }
    i++;
  }
}
int f() {
  int i, t1 = n - t, t2;
  if (t1 == 0) return 1;
  for (i = 0; i < n; i++) {
    if (s1[i] == s2[i]) {
      s3[i] = s2[i];
      t1--;
      if (t1 == 0) {
        return 1;
      }
    }
  }
  t2 = t1;
  for (i = 0; i < n; i++) {
    if (s1[i] != s2[i]) {
      if (t2 == t1) {
        s3[i] = s1[i];
        t1--;
      } else {
        s3[i] = s2[i];
        t2--;
        if (t2 == 0) return 1;
      }
    }
  }
  return -1;
}
int main() {
  cin >> n;
  cin >> t;
  cin.get();
  cin.getline(s1, n + 1);
  cin.getline(s2, n + 1);
  int x = f();
  if (x == -1)
    cout << "-1";
  else {
    complete_difer();
    for (int i = 0; i < n; i++) {
      cout << s3[i];
    }
  }
  return 0;
}