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 MOD = 1e9 + 7;
const double EPS = 1e-9;
const int MXX = 2e5 + 5;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int n;
  cin >> n;
  int k = n;
  if (n == 3)
    cout << 1 << " "
         << "" << 1 << " " << 1 << '\n';
  else {
    while (1) {
      if (n % 3 && (k - n) >= 3) break;
      n--;
    }
    cout << n << " ";
    int diff = k - n, ans1 = 0, ans2 = 0;
    for (int i = (1); i < (diff + 1); i++) {
      for (int j = (1); j < (diff + 1); j++) {
        if (i + j == diff && i % 3 != 0 && j % 3 != 0) {
          ans1 = i, ans2 = j;
          break;
        }
      }
    }
    cout << ans1 << " " << ans2 << '\n';
  }
  return 0;
}
