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.
Example
Input
5
Output
2
2 3
Input
6
Output
3
2 2 2
代码
#include <stdio.h>
#include <stdlib.h>
int main()
{
int n,j;
while(scanf("%d",&n)!=EOF)
{
if(n%2==0)
{
printf("%d
",n/2);
printf("2");
for(j=1;j<n/2;j++)
{
printf(" 2");
}
printf("
");
}
else if(n==3)
{
printf("1
3
");
}
else
{
printf("%d
",n/2);
printf("2");
for(j=2;j<n/2;j++)
{
printf(" 2");
}
printf(" 3
");
}
}
return 0;
}