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, t, diff = 0, x = 0, y = 0, mid = 0;
  scanf("%d%d", &n, &t);
  string a, b, ans = "";
  cin >> a >> b;
  for (int i = 0; i < n; ++i) {
    if (a[i] != b[i]) ++diff;
    ans += 'a';
  }
  if (diff > (2 * t)) {
    puts("-1");
    return 0;
  }
  for (int i = 0; i <= diff; ++i) {
    if ((diff - i) % 2) continue;
    if ((i + (diff - i) / 2) > t) continue;
    mid = i;
  }
  for (int i = 0; i < n; ++i) {
    if (a[i] != b[i]) {
      if (mid) {
        for (int j = 0; j < 26; ++j) {
          char c = 'a' + j;
          if ((a[i] != c) && (b[i] != c)) {
            ans[i] = c;
            ++x, ++y;
            break;
          }
        }
        --mid;
      } else {
        if (x < y)
          ++x, ans[i] = b[i];
        else
          ++y, ans[i] = a[i];
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    if (a[i] == b[i]) {
      if (x < t) {
        if (a[i] == 'a') ans[i] = 'b';
        ++x, ++y;
      } else
        ans[i] = a[i];
    }
  }
  cout << ans << endl;
  return 0;
}