B. T-primes

B. T-primes

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer t Т-prime, if t has exactly three distinct positive divisors.

You are given an array of n positive integers. For each of them determine whether it is Т-prime or not.

Input

The first line contains a single positive integer, n (1 ≤ n ≤ 10⁵), showing how many numbers are in the array. The next line contains n space-separated integers x_i (1 ≤ x_i ≤ 10¹²).

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.

Output

Print n lines: the i-th line should contain "YES" (without the quotes), if number x_i is Т-prime, and "NO" (without the quotes), if it isn't.

Sample test(s)

input

3
4 5 6

output

YES
NO
NO

Note

The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".

解题说明：此题就是判断一个数是否仅含有3个因子，其实也就是除了1和它自身外，还存在另一个因子。如果找到这个另外的因子就输出yes，否则为no。

首先来分析这个数的情况，由于只有一个不同的因子，那这个数肯定是某个数的平方数，否则不可能只有一个因子。这样问题缩小为在2到n^(1/2)-1之间求因子的个数，如果因子个数不为0，那证明还存在其他因子，不符合题意，我们就证明其中不存在因子即可，问题进一步转换为证明n^(1/2)是个素数。

证明素数的问题很常见，令m=n^(1/2)，然后判断m^(1/2)之间有没有因子即可。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;

int main()
{
	int n,i,j;
	long long a,b;
	cin>>n;
	for(i=1;i<=n;i++)
	{
		cin>>a;
		b=sqrtl(a); //long double sqrtl( long double x );
		for(j=2;j*j<=b;j++) 
		{	
			if(b%j==0)
			{
				break;
			}
		}
		if(j*j>b && b*b==a && a>1)
		{
			cout<<"YES\n";
		}
		else
		{
			cout<<"NO\n";
		}
	}
	return 0;
}