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")
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target("avx,avx2")
using namespace std;
int get_diff(string& s1, string& s2) {
  int cnt = 0;
  for (int i = 0; i < s1.size(); ++i) {
    cnt += (s1[i] != s2[i]);
  }
  return cnt;
}
void kek(string& s1, string& s2, string& s3, int& sme1, int& sme2) {
  for (int i = 0; i < s1.size(); ++i) {
    if (s3[i] != '-') continue;
    if (s1[i] == s2[i] && sme1 && sme2) {
      s3[i] = s1[i], sme1--, sme2--;
    } else if (sme1 && s1[i] != s2[i]) {
      s3[i] = s1[i], --sme1;
    }
  }
}
int main() {
  cin.tie(0), cout.tie(0), ios_base::sync_with_stdio(0);
  int n, t;
  cin >> n >> t;
  string s1, s2, s3 = "";
  cin >> s1 >> s2;
  n = s1.size();
  int d = n - t;
  for (int i = 0; i < n; ++i) s3 += '-';
  int same = max(0, 2 * d - n), sme1 = d, sme2 = d;
  int t1 = d - (n - get_diff(s1, s3)), t2 = d - (n - get_diff(s2, s3));
  for (int i = 0; i < n; ++i) {
    if (t1 && t2 && s1[i] == s2[i]) {
      --t1, --t2;
      s3[i] = s1[i];
    }
  }
  kek(s1, s2, s3, t1, t2);
  t2 = d - (n - get_diff(s2, s3));
  kek(s2, s1, s3, t2, t1);
  for (int i = 0; i < n; ++i) {
    if (s3[i] != '-') continue;
    for (char c = 'a'; c <= 'z'; ++c) {
      if (s1[i] != c && s2[i] != c) s3[i] = c;
    }
  }
  if (get_diff(s1, s3) != t || get_diff(s2, s3) != t)
    cout << -1;
  else
    cout << s3;
  return 0;
}