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;
char s1[100009], s2[100009], ans[100009];
int main() {
  cin >> n >> t;
  scanf("%s%s", s1, s2);
  int same = 0;
  int n = strlen(s1);
  for (int i = 0; i < n; i++)
    if (s1[i] == s2[i]) same++;
  int diff = n - same;
  int mi = (diff + 1) / 2;
  if (t < mi) {
    puts("-1");
  } else {
    int num;
    if (t >= diff) {
      num = diff;
    } else {
      num = t;
    }
    int pos = 0;
    for (int i = 0; i <= diff; i++) {
      if ((diff - i) % 2 == 0) {
        if ((diff - i) / 2 + i == num) {
          pos = i;
          break;
        }
      }
    }
    int cur = 0, cnt = 0;
    for (int i = 0; i < n; i++) {
      if (s1[i] != s2[i]) {
        if (cur < pos) {
          int val = 0;
          while (val + 'a' == s1[i] || val + 'a' == s2[i]) val = (val + 1) % 26;
          ans[i] = val + 'a';
          cur++;
        } else if (cnt < (diff - pos) / 2) {
          ans[i] = s1[i];
          cnt++;
        } else {
          ans[i] = s2[i];
        }
      }
    }
    cur = 0;
    for (int i = 0; i < n; i++) {
      if (s1[i] == s2[i]) {
        if (cur < t - num) {
          int val = 0;
          while (val + 'a' == s1[i] || val + 'a' == s2[i]) val = (val + 1) % 26;
          ans[i] = val + 'a';
          cur++;
        } else {
          ans[i] = s1[i];
        }
      }
    }
    ans[n] = '\0';
    puts(ans);
  }
  return 0;
}