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, k, x;
string s1, s2, s3;
int main() {
  cin >> n >> t >> s1 >> s2;
  if (t == 0) {
    if (s1 == s2)
      cout << s1;
    else
      cout << -1;
    return 0;
  }
  k = 0;
  for (int i = 0; i < n; i++)
    if (s1[i] != s2[i]) k++;
  if (k <= t) {
    x = t - k;
    s3 = "";
    for (int i = 0; i < n; i++)
      if ((s1[i] == s2[i] && x > 0) || s1[i] != s2[i]) {
        for (char c = 'a'; c <= 'z'; c++)
          if (c != s1[i] && c != s2[i]) {
            s3 += c;
            break;
          }
        if (s1[i] == s2[i]) x--;
      } else
        s3 += s1[i];
    cout << s3;
  } else {
    if (t * 2 < k) {
      cout << -1;
      return 0;
    }
    k = n - k;
    x = k = n - t - k;
    s3 = "";
    for (int i = 0; i < n; i++)
      if (s1[i] == s2[i])
        s3 += s1[i];
      else if (x > 0) {
        s3 += s1[i];
        x--;
      } else if (k > 0) {
        s3 += s2[i];
        k--;
      } else
        for (char c = 'a'; c <= 'z'; c++)
          if (c != s1[i] && c != s2[i]) {
            s3 += c;
            break;
          }
    cout << s3;
  }
  return 0;
}