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;
ostream cnull(NULL);
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cnull << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cnull << *sdbg++;
  cnull << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
const int C = 50;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int pom = n / 3;
  for (int a = max(1, pom - C); a <= min(n, pom + C); ++a) {
    for (int b = max(1, pom - C); b <= min(n, pom + C); ++b) {
      for (int c = max(1, pom - C); c <= min(n, pom + C); ++c) {
        if (a + b + c == n && a % 3 != 0 && b % 3 != 0 && c % 3 != 0) {
          cout << a << " " << b << " " << c << '\n';
          return 0;
        }
      }
    }
  }
}
