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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  string a, b;
  cin >> a >> b;
  string ans(n, '#');
  int same = n - k;
  vector<int> eqi;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      eqi.push_back(i);
    }
  }
  vector<int> done(n, 0);
  int lefta = same, leftb = same;
  for (int i : eqi) {
    if (lefta == 0) break;
    ans[i] = a[i];
    lefta--;
    leftb--;
    done[i] = 1;
  }
  for (int i = 0; i < n; i++) {
    if (done[i]) continue;
    if (lefta) {
      ans[i] = a[i];
      lefta--;
    } else if (leftb) {
      ans[i] = b[i];
      leftb--;
    } else {
      for (char c = 'a'; c <= 'z'; c++) {
        if (a[i] != c and b[i] != c) {
          ans[i] = c;
          break;
        }
      }
    }
  }
  if (lefta or leftb) {
    cout << -1;
    return 0;
  }
  cout << ans;
  return 0;
}