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 main() {
  int n, x;
  cin >> n >> x;
  x = n - x;
  string a, b, c;
  cin >> a >> b;
  vector<int> t[2];
  for (int i = 0; i < a.size(); i++) t[a[i] == b[i]].push_back(i);
  if (x > t[1].size() + t[0].size() / 2)
    cout << -1 << endl;
  else {
    c.resize(a.size());
    while (x && !t[1].empty()) {
      c[t[1].back()] = a[t[1].back()];
      t[1].pop_back();
      x--;
    }
    if (x)
      while (x && !t[0].empty()) {
        c[t[0].back()] = a[t[0].back()];
        t[0].pop_back();
        c[t[0].back()] = b[t[0].back()];
        t[0].pop_back();
        x--;
      }
    while (!t[1].empty()) {
      t[0].push_back(t[1].back());
      t[1].pop_back();
    }
    while (!t[0].empty())
      for (char x = 'a'; x <= 'c'; x++)
        if (x != a[t[0].back()] && x != b[t[0].back()]) {
          c[t[0].back()] = x;
          t[0].pop_back();
          break;
        }
    cout << c << endl;
  }
  return 0;
}