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], s[100010];
int main() {
  int n, t;
  scanf("%d%d", &n, &t);
  scanf("%s%s", s1, s2);
  queue<int> same, diff;
  for (int i = 0; i < n; i++) {
    if (s1[i] == s2[i])
      same.push(i);
    else
      diff.push(i);
  }
  t = n - t;
  if (t > same.size() && t - same.size() > diff.size() / 2)
    printf("-1\n");
  else {
    while (t > 0 && same.size() > 0) {
      int tmp = same.front();
      same.pop();
      s[tmp] = s1[tmp];
      t--;
    }
    while (t > 0 && diff.size() > 0) {
      int a = diff.front();
      diff.pop();
      int b = diff.front();
      diff.pop();
      s[a] = s1[a];
      s[b] = s2[b];
      t--;
    }
    while (same.size() > 0) {
      int tmp = same.front();
      same.pop();
      s[tmp] = (s1[tmp] - 'a' + 27) % 26 + 'a';
    }
    while (diff.size() > 0) {
      int tmp = diff.front();
      diff.pop();
      for (int i = 0; i < 26; i++) {
        if ((s1[tmp] - 'a') != i && (s2[tmp] - 'a') != i) {
          s[tmp] = i + 'a';
          break;
        }
      }
    }
    cout << s << endl;
  }
}