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;
vector<int> positions;
int n, t, cur = 0;
string a, b, ans;
int main() {
  cin >> n >> t;
  cin >> a >> b;
  for (int i = 0; i < n; ++i) {
    if (a[i] != b[i]) {
      positions.emplace_back(i);
    }
  }
  ans = a;
  int dif = positions.size();
  while (1) {
    if (t >= dif) {
      int remaining = t - dif;
      for (int j = cur; j < positions.size(); ++j) {
        int i = positions[j];
        char x;
        for (x = 'a'; x <= 'z'; ++x) {
          if (x != a[i] && x != b[i]) {
            break;
          }
        }
        ans[i] = x;
      }
      int y = 0;
      while (y < n && remaining) {
        if (a[y] == b[y]) {
          ans[y] = (a[y] == 'a' ? 'b' : 'a');
          --remaining;
        }
        ++y;
      }
      if (!remaining) {
        cout << ans << endl;
      } else {
        cout << -1;
      }
      return 0;
    } else {
      if (cur >= positions.size() - 1) {
        break;
      }
      int y = positions[cur];
      ans[y] = b[y];
      cur += 2;
      dif -= 2;
      --t;
    }
  }
  cout << "-1";
}