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>
using namespace std;
int main() {
  long long int n, a, b, c;
  cin >> n;
  a = n / 3;
  b = n / 3;
  c = n / 3 + n % 3;
  while ((a % 3 == 0) || (b % 3 == 0) || (c % 3 == 0)) {
    if (a % 3 == 0) {
      if ((b + 1) % 3 != 0) {
        b = b + 1;
        a = a - 1;
      } else if ((c + 1) % 3 != 0) {
        c++;
        a--;
      } else if ((b + 2) % 3 != 0 && a >= 3) {
        b += 2;
        a = a - 2;
      } else if ((c + 2) % 3 != 0 && a >= 3) {
        c += 2;
        a = a - 2;
      }
    }
    if (b % 3 == 0) {
      if ((a + 1) % 3 != 0) {
        a = a + 1;
        b = b - 1;
      } else if ((c + 1) % 3 != 0) {
        c++;
        b--;
      } else if ((a + 2) % 3 != 0 && b >= 3) {
        a += 2;
        b = b - 2;
      } else if ((c + 2) % 3 != 0 && b >= 3) {
        c += 2;
        b = b - 2;
      }
    }
    if (c % 3 == 0) {
      if ((b + 1) % 3 != 0) {
        b = b + 1;
        c = c - 1;
      } else if ((a + 1) % 3 != 0) {
        c--;
        a++;
      } else if ((b + 2) % 3 != 0 && c >= 3) {
        b += 2;
        c = c - 2;
      } else if ((a + 2) % 3 != 0 && c >= 3) {
        c -= 2;
        a = a + 2;
      }
    }
  }
  cout << a << " " << b << " " << c;
}
