Little C loves number «3» very much. He loves all things about it.

Now he has a positive integer n. He wants to split n into 3 positive integers a,b,c, such that a+b+c=n and none of the 3 integers is a multiple of 3. Help him to find a solution.

Input

A single line containing one integer n (3 ≤ n ≤ 10^9) — the integer Little C has.

Output

Print 3 positive integers a,b,c in a single line, such that a+b+c=n and none of them is a multiple of 3.

It can be proved that there is at least one solution. If there are multiple solutions, print any of them.

Examples

Input

3


Output

1 1 1

Input

233

Output

77 77 79
#include <bits/stdc++.h>
int main() {
  long long int n;
  scanf("%lld", &n);
  if (n % 3 == 0) {
    n = n / 3;
    if (n % 3 == 0) {
      printf("%lld %lld %lld", n - 1, n + 2, n - 1);
    } else {
      printf("%lld %lld %lld", n, n, n);
    }
  } else if ((n - 1) % 3 == 0) {
    n = (n - 1) / 3;
    if (n % 3 == 0) {
      printf("%lld %lld %lld", n - 1, n - 2, n + 4);
    } else {
      if ((n + 1) % 3 == 0) {
        printf("%lld %lld %lld", n, n - 1, n + 2);
      } else {
        printf("%lld %lld %lld", n, n, n + 1);
      }
    }
  } else if ((n - 2) % 3 == 0) {
    n = (n - 2) / 3;
    if (n % 3 == 0) {
      printf("%lld %lld %lld", n - 1, n - 1, n + 4);
    } else {
      if ((n + 2) % 3 == 0) {
        printf("%lld %lld %lld", n, n + 1, n + 1);
      } else {
        printf("%lld %lld %lld", n, n, n + 2);
      }
    }
  }
}
