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;
vector<long long> three;
bool vis[1000000 + 2];
void seive() {
  memset(vis, false, sizeof(vis));
  for (long long i = 3; i <= 1000000; i = i + 3) {
    vis[i] = true;
  }
}
int main() {
  seive();
  long long n;
  cin >> n;
  cout << "1 ";
  if (n == 3) {
    cout << "1 1";
    return 0;
  }
  n--;
  for (long long j = n - 1; j > 1; j--) {
    if (j % 3 != 0) {
      long long y = n - j;
      if (vis[y] == false) {
        cout << y << " " << j;
        break;
      }
    }
  }
}
