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, d;
char a[1000006], b[1000005], ans[100005];
char none(char a, char b) {
  if (a != 'a' && b != 'a') return 'a';
  if (a != 'b' && b != 'b') return 'b';
  return 'c';
}
int main() {
  scanf("%d%d", &n, &t);
  scanf(" %s%s", a, b);
  for (int i = 0; i < n; i++) {
    if (a[i] != b[i]) d++;
  }
  int e = n - d;
  int mx = n, mn = d / 2;
  int d0 = d % 2 == 1, t0 = 0, t1 = 0;
  if (t - d0 > n - d0 || t - d0 < mn) {
    printf("-1\n");
    return 0;
  }
  t0 = t1 = t;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) continue;
    if (--d0 == 0) {
      d--;
      ans[i] = none(a[i], b[i]);
      t--;
      t0--;
      t1--;
      continue;
    }
    if (t1 > t0) {
      ans[i] = a[i];
      t1--;
      t--;
      d--;
      continue;
    } else if (t0 > t1) {
      ans[i] = b[i];
      t0--;
      t--;
      d--;
      continue;
    }
    if (d + e > t) {
      ans[i] = b[i];
      t0--;
      d--;
      continue;
    } else {
      ans[i] = none(b[i], a[i]);
      t--;
      t0--;
      t1--;
      d--;
      continue;
    }
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      if (t) {
        ans[i] = none(b[i], a[i]);
        t--;
      } else {
        ans[i] = a[i];
      }
    }
  }
  printf("%s\n", ans);
  return 0;
}