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】:

【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 
     
      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
      
        pii;typedef pair
       
         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 
        
          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;}