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;
template <class U, class T>
void Max(U &first, T second) {
  if (first < second) first = second;
}
template <class U, class T>
void Min(U &first, T second) {
  if (first > second) first = second;
}
template <class T>
void add(int &a, T b) {
  a = (a + b) % 1000000007;
}
inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % 1000000007;
    a = 1LL * a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
int pow(int a, int b, int c) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % c;
    a = 1LL * a * a % c;
    b >>= 1;
  }
  return ans;
}
int main() {
  int ca = 0, T, k, i, j, m = 0, K, n;
  scanf("%d", &n);
  for (int i = 1; i < 3; i++) {
    for (int j = 1; j < 3; j++) {
      for (int k = 1; k < 3; k++) {
        m = i + j + k;
        if (m > n) continue;
        m = n - m;
        if (m % 3) continue;
        m /= 3;
        printf("%d %d %d\n", i, j, 3 * m + k);
        return 0;
      }
    }
  }
}
