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 s1[100010], s2[100010], s3[100010];
int main() {
  int a, b;
  bool flag;
  int i, n, t;
  char ch;
  int ns1, ns2;
  while (scanf("%d%d", &n, &t) != EOF) {
    scanf("%s%s", s1, s2);
    a = b = 0;
    flag = true;
    for (i = 0; i < n; i++) {
      a += (s1[i] == s2[i]);
    }
    if (n - t <= a) {
      ns1 = n - t;
      for (i = 0; i < n; i++)
        if (s1[i] == s2[i] && ns1 > 0) {
          s3[i] = s1[i];
          ns1--;
        } else
          for (ch = 'a'; ch <= 'z'; ch++)
            if (ch != s1[i] && ch != s2[i]) {
              s3[i] = ch;
              break;
            }
    } else {
      ns1 = ns2 = n - t - a;
      for (i = 0; i < n; i++)
        if (s1[i] == s2[i])
          s3[i] = s1[i];
        else {
          if (ns1 > 0) {
            s3[i] = s1[i];
            ns1--;
          } else if (ns2 > 0) {
            s3[i] = s2[i];
            ns2--;
          } else {
            for (ch = 'a'; ch <= 'z'; ch++)
              if (ch != s1[i] && ch != s2[i]) {
                s3[i] = ch;
                break;
              }
          }
        }
      if (ns1 > 0 || ns2 > 0) flag = false;
    }
    s3[n] = '\0';
    if (flag)
      printf("%s\n", s3);
    else
      printf("-1\n");
  }
  return 0;
}