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(void) {
  string s1, s2, s3;
  int a, b;
  cin >> a >> b >> s1 >> s2;
  s3 = s1;
  int sim = 0, disim = 0;
  for (int i = 0; i < a; i++) {
    if (s1[i] == s2[i])
      sim++;
    else
      disim++;
  }
  if (disim > 2 * b) {
    cout << "-1" << endl;
  } else {
    if (disim <= b) {
      for (int i = 0; i < a; i++) {
        if (s1[i] != s2[i]) {
          for (int j = 'a'; j <= 'z'; j++) {
            if (j != s1[i] && j != s2[i]) {
              s3[i] = j;
              b--;
              break;
            }
          }
        }
      }
      for (int i = 0; i < a && b; i++) {
        if (s1[i] == s2[i]) {
          if (s1[i] == 'z')
            s3[i] = 'a';
          else
            s3[i] = s1[i] + 1;
          b--;
        }
      }
    } else {
      int temp = disim - ((disim - b) * 2);
      int k = b - temp;
      int i = 0;
      for (; i < a && temp; i++) {
        if (s1[i] != s2[i]) {
          for (int j = 'a'; j <= 'z'; j++) {
            if (j != s1[i] && j != s2[i]) {
              s3[i] = j;
              temp--;
              break;
            }
          }
        }
      }
      int flag = 0;
      for (int j = 0; i < a && k; i++) {
        if (s1[i] != s2[i]) {
          if (flag) {
            s3[i] = s1[i];
            flag = 0;
            j++;
          } else {
            s3[i] = s2[i];
            flag = 1;
            j++;
          }
          if (j % 2 == 0 && j != 0) k--;
        }
      }
    }
    cout << s3 << endl;
  }
  return 0;
}