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;
const int N = 1e5 + 5;
int n, v[N];
void Print() {
  for (int i = 1; i <= n; i++) cout << char(v[i] + 'a');
  exit(0);
}
int main() {
  int need;
  cin >> n >> need;
  string a, b;
  cin >> a >> b;
  a = "x" + a;
  b = "x" + b;
  for (int i = 1; i <= n; i++) v[i] = -1;
  need = n - need;
  for (int i = 1; i <= n; i++) {
    if (need == 0) break;
    if (a[i] == b[i]) {
      v[i] = a[i] - 'a';
      need--;
    }
  }
  if (need == 0) {
    for (int i = 1; i <= n; i++) {
      if (v[i] != -1) continue;
      v[i] = 0;
      while (v[i] == a[i] - 'a' || v[i] == b[i] - 'a') v[i]++;
    }
    Print();
  }
  int have = 0;
  for (int i = 1; i <= n; i++)
    if (v[i] == -1) have++;
  if (2 * need > have) {
    cout << "-1\n";
    return 0;
  }
  int F = need, S = need;
  for (int i = 1; i <= n; i++) {
    if (v[i] != -1) continue;
    if (F) {
      v[i] = a[i] - 'a';
      F--;
      continue;
    }
    if (S == 0) break;
    v[i] = b[i] - 'a';
    S--;
  }
  for (int i = 1; i <= n; i++) {
    if (v[i] != -1) continue;
    v[i] = 0;
    while (v[i] == a[i] - 'a' || v[i] == b[i] - 'a') v[i]++;
  }
  Print();
  return 0;
}