CF230b： T-primes（简单数论）

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 tТ-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 nspace-separated integers xi (1 ≤ xi ≤ 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 %I64dspecifier.

Output

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

Examples

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".

题意：判断一个数是否只有3个因子

思路：该数一定是质数的平方。

# include <stdio.h>
# include <math.h>
# define MAXN 1000000
int p[MAXN+1]={1,1};
int main()
{
    int i, j, t;
    long long n;
    double temp;
    for(i=2; i<=MAXN; ++i)
        if(!p[i])
        for(j=i+i; j<=MAXN; j+=i)
            p[j] = 1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d", &n);
        temp = sqrt(n);
        if(temp == (int)temp && !p[(int)temp])
            puts("YES");
        else
            puts("NO");

    }
    return 0;
}