We call a positive number
X
P-number if there is not a positive number that is less than
X
and the greatest common divisor of these two numbers is bigger than 1.
Now you are given a sequence of integers. You task is to calculate the sum of P-numbers of the sequence.

There are several test cases.
In each test case:
The first line contains a integer N(1≤N≤1000).
The second line contains N
integers. Each integer is between 1 and 1000.

For each test case, output the sum of P-numbers of the sequence.

3
5 6 7
1
10

12
0
/*
HDU 5150 素数求和 
注意1也算。。。 
*/
#include<iostream>
#include<stdio.h>
using namespace std;
int hash[1001];

void getResult()
{
	int i,j;
	memset(hash,0,sizeof(hash));
	hash[1]=1;
	for(i=2;i<1001;i++)
	{
		if(hash[i]==0)
		{
			hash[i]=1;
			for(j=i*2;j<1001;j+=i)
				hash[j]=-1;
		}
	}
	
	
}
int main()
{
	int n,a,sum,i;
	getResult();
	while(scanf("%d",&n)!=EOF)
	{
		sum=0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&a);
			if(hash[a]==1)
				sum+=a;
		}
		printf("%d\n",sum);	
	}
	return 0;
}