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;
long long int n, t;
string s1, s2;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> t >> s1 >> s2;
  long long int same = 0, diff = 0;
  for (long long int i = 0; i < (long long int)s1.length(); ++i) {
    if (s1[i] == s2[i])
      same++;
    else
      diff++;
  }
  if (n - t > same + diff / 2) {
    cout << "-1\n";
    return 0;
  }
  if (n - t < same) {
    long long int d = n - t;
    for (long long int i = 0; i < n; ++i) {
      if (s1[i] == s2[i] && d) {
        d--;
        cout << s1[i];
      } else {
        for (char a = 'a'; a <= 'z'; ++a) {
          if (a != s1[i] && a != s2[i]) {
            cout << a;
            break;
          }
        }
      }
    }
    return 0;
  }
  long long int d = (n - t) - same;
  long long int prev = 0;
  for (long long int i = 0; i < n; ++i) {
    if (!d && (s1[i] != s2[i])) {
      for (char a = 'a'; a <= 'z'; ++a) {
        if (a != s1[i] && a != s2[i]) {
          cout << a;
          break;
        }
      }
    } else if (s1[i] == s2[i]) {
      cout << s1[i];
    } else {
      if (prev % 2 == 0)
        cout << s1[i];
      else
        cout << s2[i];
      prev++;
      if (prev % 2 == 0) d--;
    }
  }
  cout << '\n';
  return 0;
}