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;
struct P {
  int x;
  long long y;
  bool operator<(const P &a) const { return y > a.y; }
};
vector<long long> v;
int a, c, i, b, n, m, k, d;
long long o[302022];
int l[1];
long long j[45];
int e[101111];
int dx[10] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[10] = {1, 0, -1, 0, 1, -1, 1, -1},
    dz[10] = {0, 0, 0, 0, 1, -1};
long long x, y, mod = 998244353, mod2 = 1000000009, mod3 = 2017;
long long z;
double pi = 3.14159265;
queue<int> q;
char r[333333];
bool as(P a, P b) {
  if (a.x != b.x) return a.x < b.x;
  return a.y < b.y;
}
int f(char a) {
  if (a <= 'Z') return a - 'A';
  return a - 'a' + 26;
}
int main() {
  scanf("%d", &n);
  a = n / 3;
  b = n / 3;
  if (a % 3 == 0) a--, b--;
  c = n - a - b;
  if (c % 3 == 0) c -= b % 3, b += b % 3;
  printf("%d %d %d", a, b, c);
}
