1047A_Little C Loves 3 I（构造）

A. Little C Loves 3 I

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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

Input

A single line containing one integer $\mathrm{nn (3\le n\le 1093\le n\le 109)\; —\; the\; integer\; Little\; C\; has.}$

Output

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

It can be proved that there is at least one solution. If there are multiple solutions, print any of them.

Examples

input

Copy

3

output

Copy

1 1 1

input

Copy

233

output

Copy

77 77 79
被样例困惑了很久，看了题解啧啧，其实这题就是构造
大致题意：给你一个数n,找到三个数a,b,c，使得a+b+c=n,并且a,b,c都不能是3的倍数
分析：当n%3=0时，n-2一定不是3的倍数，可以构造为a=1,b=1,c=n-2;当n%3!=0时，n-3一定不是3的倍数，那么可以构造为，a=1,b=2,c=n-3.注意这个构造是任意的，只要满足条件即可

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n%3==0)
            printf("1 1 %d\n",n-2);
        else
            printf("1 2 %d\n",n-3);
    }
    return 0;
}