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 <typename T>
void _dbg(char const *s, const T &t) {
  cout << s << "=" << t << endl;
}
template <typename T, typename... TA>
void _dbg(char const *s, const T &t, const TA &...ta) {
  while (*s != ',') cout << *s++;
  cout << "=" << t << ",";
  _dbg(s + 1, ta...);
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  return os << "{" << p.first << "," << p.second << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ",";
    os << *it;
  }
  return os << "}";
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const map<T1, T2> &mp) {
  os << "{";
  for (auto it = mp.begin(); it != mp.end(); ++it) {
    if (it != mp.begin()) os << ",";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &s) {
  os << "{";
  for (auto it = s.begin(); it != s.end(); ++it) {
    if (it != s.begin()) os << ",";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &s) {
  os << "{";
  for (auto it = s.begin(); it != s.end(); ++it) {
    if (it != s.begin()) os << ",";
    os << *it;
  }
  return os << "}";
}
template <size_t N>
ostream &operator<<(ostream &os, const bitset<N> &bs) {
  os << "{";
  for (size_t i = 0; i < N; ++i) {
    cout << bs[i];
  }
  return os << "}";
}
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<ld, ld>;
using pil = pair<int, ll>;
void solve() {
  int n;
  scanf("%d", &n);
  for (int a = 1; a <= 10; ++a) {
    for (int b = 1; b <= 10; ++b) {
      int c = n - a - b;
      if (c <= 0) continue;
      if (a % 3 == 0 || b % 3 == 0 || c % 3 == 0) continue;
      printf("%d %d %d\n", a, b, c);
      return;
    }
  }
}
int main() {
  solve();
  return 0;
}
