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;
int n, t, s, d, c;
string s1, s2;
vector<int> same, diff;
vector<char> ans(1e5 + 5);
char get(char a, char b) {
  for (int i = 97; i <= 122; ++i) {
    if (char(i) != a && char(i) != b) {
      return char(i);
    }
  }
}
int main(void) {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> t >> s1 >> s2;
  s1 = '*' + s1;
  s2 = '*' + s2;
  for (int i = 1; i <= n; ++i) {
    if (s1[i] == s2[i]) {
      same.emplace_back(i);
    } else {
      diff.emplace_back(i);
    }
  }
  s = int(same.size());
  d = int(diff.size());
  if (t < (d / 2 + (d & 1 ? 1 : 0))) {
    cout << -1 << "\n";
    return 0;
  }
  c = max(0, t - d);
  for (int i = 0; i <= c - 1; ++i) {
    --t;
    ans[same[i]] = get(s1[same[i]], s1[same[i]]);
  }
  for (int i = c; i <= s - 1; ++i) {
    ans[same[i]] = s1[same[i]];
  }
  for (int i = 0; i <= 2 * (d - t) - 1; i += 2) {
    ans[diff[i]] = s1[diff[i]];
    ans[diff[i + 1]] = s2[diff[i + 1]];
  }
  for (int i = 2 * (d - t); i <= d - 1; ++i) {
    ans[diff[i]] = get(s1[diff[i]], s2[diff[i]]);
  }
  cout << string(ans.begin() + 1, ans.begin() + n + 1) << "\n";
  return 0;
}