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() {
  long n, t;
  char s1[100001], s2[100001], answer[100001];
  cin >> n >> t;
  n--;
  cin >> s1 >> s2;
  long dif = 0;
  for (long i = 0; i <= n; i++)
    if (s1[i] != s2[i]) dif++;
  if (dif > t) {
    long n1 = dif - t;
    long n2 = n1;
    if (n1 + n2 > dif) {
      cout << "-1\n";
      return 0;
    }
    for (long i = 0; i <= n; i++) {
      if (s1[i] == s2[i])
        answer[i] = s1[i];
      else {
        if (n1) {
          answer[i] = s1[i];
          n1--;
        } else if (n2) {
          answer[i] = s2[i];
          n2--;
        } else {
          char c;
          for (long j = 0; j <= 25; j++) {
            c = 'a' + j;
            if (c != s1[i] && c != s2[i]) break;
          }
          answer[i] = c;
        }
      }
    }
  } else {
    long NN = t - dif;
    for (long i = 0; i <= n; i++) {
      if (s1[i] == s2[i]) {
        if (NN) {
          if (s1[i] == 'z')
            answer[i] = 'x';
          else
            answer[i] = s1[i] + 1;
          NN--;
        } else
          answer[i] = s1[i];
      } else {
        char c;
        for (long j = 0; j <= 25; j++) {
          c = 'a' + j;
          if (c != s1[i] && c != s2[i]) break;
        }
        answer[i] = c;
      }
    }
  }
  cout << answer;
  return 0;
}