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>
#pragma GCC optimize("O3")
using namespace std;
const long long MAX = 25e4 + 5;
const long long MAX2 = 11;
const int MOD = 1000000000 + 7;
const long long INF = 20000;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
int n, m, a, b;
string s, t, ans;
bool st;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  m = n - m;
  cin >> s >> t;
  n--;
  for (long long i = 0; i <= n; i++) {
    if (s[i] == t[i])
      a++;
    else
      b++;
  }
  b /= 2;
  if (a + b < m) return cout << -1, 0;
  for (long long i = 0; i <= n; i++) {
    if (s[i] == t[i]) {
      if (!m) {
        if (s[i] == 'a')
          ans += 'b';
        else
          ans += 'a';
      } else
        m--, ans += s[i];
    } else {
      if (st)
        ans += t[i], st = 0;
      else if (!m || !b) {
        for (long long j = 'a'; j <= 'z'; j++)
          if (j != s[i] && j != t[i]) {
            ans += j;
            break;
          }
      } else
        b--, m--, ans += s[i], st = 1;
    }
  }
  cout << ans << endl;
  return 0;
}