Problem: 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 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 %I64d specifier.

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".

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
string to_str(T str) {
  stringstream stream;
  stream << str;
  return stream.str();
}
template <typename T>
int to_int(T num) {
  int val;
  stringstream stream;
  stream << num;
  stream >> val;
  return val;
}
vector<string> split(const string& s, char delim) {
  vector<string> elems;
  stringstream ss(s);
  string item;
  while (getline(ss, item, delim)) elems.push_back(item);
  return elems;
}
bool comp(const pair<int, int>& i, const pair<int, int>& j) {
  return i.second < j.second;
}
long int setbit(long int n, long int pos) { return n = n | (1 << pos); }
long int resbit(long int n, long int pos) { return n = n & ~(1 << pos); }
bool checkbit(long long int n, long long int pos) {
  return (bool)(n & (1ll << pos));
}
long long int bigmod(long long int b, long long int p, long long int m) {
  if (p == 0) return 1;
  long long int ret = bigmod(b, p / 2, m);
  ret = (ret * ret) % m;
  if (p & 1) ret = (ret * b) % m;
  return ret;
}
const int dr[] = {-1, -1, 0, 1, 1, 1, 0, -1};
const int dc[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int T2 = 105;
const int T3 = 1005;
const int T4 = 10005;
const int T5 = 100005;
int i = 0, j = 0, k = 0, t = 0, n;
int prime[1000005];
void pre() {
  prime[0] = 1, prime[1] = 1;
  for (int i = 2; i * i < 1000005; i++) {
    for (int j = i * i; j < 1000005; j = j + i) prime[j] = 1;
  }
}
int main() {
  pre();
  scanf("%d", &n);
  long long int x, t;
  for (int i = 0; i < (n); i++) {
    cin >> x;
    t = sqrt(x);
    if (prime[t] == 0 && t * t == x)
      printf("YES\n");
    else
      printf("NO\n");
  }
}