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;
vector<int> mil, omil;
int main() {
  int n, t;
  string a, b;
  char c[100010];
  cin >> n >> t;
  cin >> a;
  cin >> b;
  t = n - t;
  for (int i = 0; i < a.size(); i++) {
    if (a[i] == b[i])
      mil.push_back(i);
    else
      omil.push_back(i);
  }
  if (mil.size() < t) {
    t -= mil.size();
    if (omil.size() < (2 * t)) {
      puts("-1");
      return 0;
    }
    for (int i = 0; i < mil.size(); i++) c[mil[i]] = a[mil[i]];
    for (int i = 0; i < t; i++) c[omil[i]] = a[omil[i]];
    for (int i = t; i < 2 * t; i++) c[omil[i]] = b[omil[i]];
    for (int i = 2 * t; i < omil.size(); i++) {
      for (char ch = 'a'; ch <= 'z'; ch++) {
        if (ch != a[omil[i]] && ch != b[omil[i]]) {
          c[omil[i]] = ch;
          break;
        }
      }
    }
    puts(c);
    return 0;
  }
  for (int i = 0; i < t; i++) c[mil[i]] = a[mil[i]];
  for (int i = t; i < mil.size(); i++) {
    for (char ch = 'a'; ch <= 'z'; ch++) {
      if (ch != a[mil[i]] && ch != b[mil[i]]) {
        c[mil[i]] = ch;
        break;
      }
    }
  }
  for (int i = 0; i < omil.size(); i++) {
    for (char ch = 'a'; ch <= 'z'; ch++) {
      if (ch != a[omil[i]] && ch != b[omil[i]]) {
        c[omil[i]] = ch;
        break;
      }
    }
  }
  puts(c);
}