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 INF = 0x3f3f3f3f;
const double EPS = 1e-8;
const int MOD = 1000000007;
const int N = 2000000 + 10;
const int M = 2000000 + 10;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  int tmp = n / 3;
  if (n % 3 == 0) {
    if (tmp % 3)
      cout << tmp << ' ' << tmp << ' ' << tmp << '\n';
    else
      cout << tmp - 2 << ' ' << tmp + 1 << ' ' << tmp + 1 << '\n';
  } else if (n % 3 == 1) {
    if (tmp % 3 == 1)
      cout << tmp + 1 << ' ' << tmp << ' ' << tmp << '\n';
    else if (tmp % 3 == 2)
      cout << tmp + 2 << ' ' << tmp << ' ' << tmp - 1 << '\n';
    else
      cout << tmp + 4 << ' ' << tmp - 2 << ' ' << tmp - 1 << '\n';
  } else {
    if (tmp % 3 == 1)
      cout << tmp + 1 << ' ' << tmp + 1 << ' ' << tmp << '\n';
    else if (tmp % 3 == 2)
      cout << tmp + 2 << ' ' << tmp << ' ' << tmp << '\n';
    else
      cout << tmp + 2 << ' ' << tmp + 1 << ' ' << tmp - 1 << '\n';
  }
  return 0;
}
