codeforces—— 230B —— 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 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

34 5 6

output

YESNONO

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".

问一个数是不是完全平方数。（注意数据大小）

#include<iostream>#include<cmath>#include<cstdio>#include<map>#include<set>#include<cstring>#include<string>#include<algorithm>#define min 1e-6using namespace std;#define PrimeMax 1000000bool Prime[PrimeMax+1];void IsPrime(){    memset(Prime,true,sizeof(Prime));    Prime[1]=Prime[0]=false;    int n=PrimeMax,m=sqrt(PrimeMax);    for(int i=2; i<=m; i++)        if(Prime[i])            for(int j=i*i; j<=n; j+=i)                Prime[j]=false;}int main(){    IsPrime();    int N;    cin>>N;    while(N--)    {        long long n;        cin>>n;        long long m=sqrt(n);        if((m*m==n)&&Prime[m])            cout<<"YES"<<endl;        else            cout<<"NO"<<endl;    }    return 0;}