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;
char A[100005], B[100005], C[100005];
int main() {
  int c1, i, j, k, m, n, t, count, q, c, c_a, c_b;
  char ch;
  cin >> n;
  cin >> t;
  for (i = 0; i < n; i++) C[i] = '0';
  scanf(" %s", A);
  scanf(" %s", B);
  q = 0;
  for (i = 0; i < n; i++) {
    if (A[i] == B[i]) q++;
  }
  k = n - t;
  if (k <= q) {
    c1 = 0;
    for (i = 0; i < n; i++) {
      if (A[i] == B[i] && c1 < k) {
        C[i] = A[i];
        c1++;
      } else {
        ch = A[i];
        while (1) {
          ch++;
          if (ch > 'z') ch = 'a';
          if (ch != A[i] && ch != B[i]) break;
        }
        C[i] = ch;
      }
    }
    for (i = 0; i < n; i++) cout << C[i];
    cout << endl;
  } else {
    if (n - q < 2 * (k - q))
      cout << "-1" << endl;
    else {
      for (i = 0; i < n; i++) {
        if (A[i] == B[i]) C[i] = A[i];
      }
      c_a = 0;
      c_b = 0;
      for (i = 0; i < n; i++) {
        if (C[i] == '0') {
          if (c_a < k - q) {
            C[i] = A[i];
            c_a++;
          } else if (c_b < k - q) {
            C[i] = B[i];
            c_b++;
          } else {
            ch = A[i];
            while (1) {
              ch++;
              if (ch > 'z') ch = 'a';
              if (ch != A[i] && ch != B[i]) break;
            }
            C[i] = ch;
          }
        }
      }
      for (i = 0; i < n; i++) cout << C[i];
      cout << endl;
    }
  }
}