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;
const int N = 100005, M = 1000000007, BIG = 0x3f3f3f3f;
int n;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  while (cin >> n) {
    int rem = n % 3;
    int q = n / 3;
    if (rem == 1) {
      if (q % 3) {
        if ((q + 1) % 3) {
          cout << q << " " << q << " " << q + 1 << "\n";
        } else {
          cout << q << " " << q - 1 << " " << q + 2 << "\n";
        }
      } else {
        cout << q - 1 << " " << q + 1 << " " << q + 1 << "\n";
      }
    } else if (rem == 2) {
      if (q % 3) {
        if ((q + 1) % 3) {
          cout << q << " " << q + 1 << " " << q + 1 << "\n";
        } else {
          cout << q << " " << q << " " << q + 2 << "\n";
        }
      } else {
        cout << q - 1 << " " << q + 1 << " " << q + 2 << "\n";
      }
    } else {
      if (q % 3)
        cout << q << " " << q << " " << q << "\n";
      else
        cout << q - 1 << " " << q - 1 << " " << q + 2 << "\n";
    }
    if (cin.eof()) break;
  }
  return 0;
}
