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>
namespace chtholly {
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) f ^= c == '-';
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ '0');
  return f ? x : -x;
}
template <typename mitsuha>
inline bool read(mitsuha &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c) && ~c; c = getchar()) f ^= c == '-';
  if (!~c) return 0;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ '0');
  return x = f ? x : -x, 1;
}
template <typename mitsuha>
inline int write(mitsuha x) {
  if (!x) return 0 & putchar(48);
  if (x < 0) x = -x, putchar('-');
  int bit[20], i, p = 0;
  for (; x; x /= 10) bit[++p] = x % 10;
  for (i = p; i; --i) putchar(bit[i] + 48);
  return 0;
}
inline char fuhao() {
  char c = getchar();
  for (; isspace(c); c = getchar())
    ;
  return c;
}
}  // namespace chtholly
using namespace chtholly;
using namespace std;
int main() {
  int n = read();
  if (n % 3 == 0)
    printf("%d %d %d\n", 1, 1, n - 2);
  else if (n % 3 == 1)
    printf("%d %d %d\n", 1, 2, n - 3);
  else
    printf("%d %d %d\n", 2, 2, n - 4);
}
