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 main() {
  int n, t, c = 0;
  bool res = true;
  cin >> n >> t;
  int t1 = t, t2 = t;
  string s1, s2, s3;
  cin >> s1 >> s2;
  s3 = s1;
  for (int i = 0; i < n; ++i) {
    if (s1[i] == s2[i]) c++;
  }
  for (int i = 0; i < n; ++i) {
    if (s1[i] != s2[i]) {
      if (t1 == 0 && t2 == 0) {
        cout << -1;
        return 0;
      } else if (t1 == 0) {
        s3[i] = s1[i];
        t2--;
      } else if (t2 == 0) {
        s3[i] = s2[i];
        t1--;
      } else {
        s3[i] = s1[i];
        t2--;
      }
    }
  }
  if (t1 > 0) {
    for (int i = 0; i < n; ++i) {
      if (s1[i] != s2[i] && s3[i] == s1[i] && t1 > 0) {
        t1--;
        if (s1[i] != 'a' && s2[i] != 'a')
          s3[i] = 'a';
        else if (s1[i] != 'b' && s2[i] != 'b')
          s3[i] = 'b';
        else
          s3[i] = 'c';
      }
    }
  }
  if (t2 > 0) {
    for (int i = 0; i < n; ++i) {
      if (s1[i] != s2[i] && s3[i] == s2[i] && t2 > 0) {
        t2--;
        if (s1[i] != 'a' && s2[i] != 'a')
          s3[i] = 'a';
        else if (s1[i] != 'b' && s2[i] != 'b')
          s3[i] = 'b';
        else
          s3[i] = 'c';
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    if (s1[i] == s2[i]) {
      if (t1 > 0 && t2 > 0) {
        s3[i] = (s1[i] == 'a') ? 'b' : 'a';
        t1--;
        t2--;
      } else
        s3[i] = s1[i];
    }
  }
  if (t1 > 0 || t2 > 0)
    cout << -1;
  else
    cout << s3;
  return 0;
}