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>
int main() {
  int num;
  scanf("%d", &num);
  int t1, t2;
  t1 = num / 3;
  t2 = num % 3;
  if (t2 == 0) {
    if (t1 % 3 == 0) {
      printf("%d %d %d", t1 + 1, t1 - 2, t1 + 1);
      return 0;
    } else {
      printf("%d %d %d", t1, t1, t1);
      return 0;
    }
  } else {
    if (t1 % 3 == 0) {
      if (t2 == 1) {
        printf("%d %d %d", t1 - 1, t1 + 1, t1 + 1);
        return 0;
      } else if (t2 == 2) {
        printf("%d %d %d", t1 - 1, t1 + 2, t1 + 1);
        return 0;
      }
    }
    if (t1 % 3 != 0) {
      if ((t1 + 1) % 3 != 0) {
        if (t2 == 1) {
          printf("%d %d %d", t1, t1, t1 + 1);
          return 0;
        } else if (t2 == 2) {
          printf("%d %d %d", t1, t1 + 1, t1 + 1);
          return 0;
        }
      } else if ((t1 + 2) % 3 != 0) {
        if (t2 == 1) {
          printf("%d %d %d", t1, t1 + 2, t1 - 1);
          return 0;
        } else if (t2 == 2) {
          printf("%d %d %d", t1, t1, t1 + 2);
          return 0;
        }
      }
    }
  }
  return 0;
}
