time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.

Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.

Input

The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).

Output

The first line of the output contains a single integer k — maximum possible number of primes in representation.

The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.

Examples

input

5

output

2

2 3

input

6

output

3

2 2 2

【题目链接】:http://codeforces.com/contest/749/problem/A

【题解】



显然奇数的话就先减个3，然后就是偶数了，一直减2就好;

偶数的话就全都是2。

我写了个枚举。

不知道自己怎么想的。



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define rei(x) scanf("%d",&x)

#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

//const int MAXN = x;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

int n;

vector <int> a,ans;

bool is(int x)

{

    int len = sqrt(x);

    rep1(i,2,len)

        if ((x%i)==0)

            return false;

    return true;

}

int main()

{

    //freopen("F:\\rush.txt","r",stdin);

    rei(n);

    rep1(i,2,n)

        if (is(i))

            a.pb(i);

    int now = 0;

    while (n)

    {

        if (n-a[now]>1 || n-a[now]==0)

        {

            n-=a[now];

            ans.pb(a[now]);

        }

        else

        {

            now++;

            ans.pb(a[now]);

            n-=a[now];

        }

    }

    int len = ans.size();

    printf("%d\n",len);

    rep1(i,0,len-1)

    {

        printf("%d",ans[i]);

        if (i==len-1)

            puts("");

        else

            putchar(' ');

    }

    return 0;

}