Little C loves number «3» very much. He loves all things about it.

Now he has a positive integer nn. He wants to split nn into 33 positive integers a,b,ca,b,c, such that a+b+c=na+b+c=n and none of the 33 integers is a multiple of 33. Help him to find a solution.

Input

A single line containing one integer nn (3≤n≤1093≤n≤109) — the integer Little C has.

Output

Print 33 positive integers a,b,ca,b,c in a single line, such that a+b+c=na+b+c=n and none of them is a multiple of 33.

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

把一個數拆成三個數，並且使每個數都是3的倍數

直接分類討論即可

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define ll long long
#define ull unsigend long long
const int M=1e3+5;
const int N=998244353;
int n,a,b,c;
void solve(int n)
{
   if(n%3==0)
   {
       printf("%d %d %d\n",1,1,n-2);
   }
   else if(n%3==1)
   {
       printf("%d %d %d\n",1,1,n-2);
   }
   else if(n%3==2)
   {
       printf("%d %d %d\n",1,2,n-3);
   }
}
int main()
{
    int n,m;
    while(~scanf("%d",&n))
    {
        solve(n);
    }
    return 0;
}