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 n, t, m;
char s1[100005], s2[100005], s3[100005];
int main() {
  int i, j, k;
  scanf("%d%d", &n, &t);
  scanf("%s", s1);
  scanf("%s", s2);
  for (i = 0, m = 0; i < n; i++)
    if (s1[i] != s2[i]) m++;
  if ((m + 1) / 2 > t)
    printf("-1\n");
  else {
    memset(s3, 0, sizeof(s3));
    t = n - t;
    if (n - m >= t) {
      for (i = j = 0; i < n; i++) {
        if (s1[i] == s2[i] && j < t)
          s3[i] = s1[i], j++;
        else {
          if (s1[i] == 'a' || s2[i] == 'a') {
            if (s1[i] == 'b' || s2[i] == 'b')
              s3[i] = 'c';
            else
              s3[i] = 'b';
          } else
            s3[i] = 'a';
        }
      }
    } else {
      for (i = 0, j = k = t - (n - m); i < n; i++) {
        if (s1[i] == s2[i])
          s3[i] = s1[i];
        else if (j > 0)
          s3[i] = s1[i], j--;
        else if (k > 0)
          s3[i] = s2[i], k--;
        else {
          if (s1[i] == 'a' || s2[i] == 'a') {
            if (s1[i] == 'b' || s2[i] == 'b')
              s3[i] = 'c';
            else
              s3[i] = 'b';
          } else
            s3[i] = 'a';
        }
      }
    }
    printf("%s\n", s3);
  }
  return 0;
}