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>
const int NMAX = 1000 * 1000 + 7;
char s1[NMAX], s2[NMAX], s[NMAX];
using namespace std;
char my_search(int i) {
  if (s1[i] != 'a' && s2[i] != 'a')
    return 'a';
  else if (s1[i] != 'b' && s2[i] != 'b')
    return 'b';
  else if (s1[i] != 'c' && s2[i] != 'c')
    return 'c';
  else if (s1[i] != 'd' && s2[i] != 'd')
    return 'd';
}
int main() {
  int n, t;
  cin >> n >> t;
  t = n - t;
  for (int i = 0; i < n; ++i) cin >> s1[i];
  for (int i = 0; i < n; ++i) cin >> s2[i];
  int d = 0;
  for (int i = 0; i < n; ++i) d += (s1[i] == s2[i]);
  if (d < t) {
    if ((t - d) * 2 > (n - d)) {
      cout << "-1";
      return 0;
    }
    int a = 0, b = 0;
    for (int i = 0; i < n; ++i)
      if (s1[i] == s2[i])
        s[i] = s1[i];
      else {
        if (a < t - d) {
          a++;
          s[i] = s1[i];
        } else if (b < t - d) {
          b++;
          s[i] = s2[i];
        } else
          s[i] = my_search(i);
      }
  } else if (d == t) {
    for (int i = 0; i < n; ++i)
      if (s1[i] == s2[i])
        s[i] = s1[i];
      else
        s[i] = my_search(i);
  } else if (d > t) {
    int a = 0;
    for (int i = 0; i < n; ++i)
      if (s1[i] == s2[i] && a < t) {
        s[i] = s1[i];
        a++;
      } else
        s[i] = my_search(i);
  }
  for (int i = 0; i < n; ++i) cout << s[i];
  return 0;
}