Gym 100796K Profact

Alice is bored out of her mind by her math classes. She craves for something much more exciting. That is why she invented a new type of numbers, the profacts. Alice calls a positive integer number a profact if it can be
expressed as a product of one or several factorials.

Just today Alice received n bills. She wonders whether the costs on the bills are profact numbers. But the numbers are too large, help Alice check this!

Input

The first line contains a single integer n, the number of bills (1 ≤ n ≤ 105). Each of the next n lines
contains a single integer ai, the cost on the i-th bill (1 ≤ ai ≤ 1018).

Output

Output n lines, on the i-th line output the answer for the number ai.
If the number ai is a profact, output "YES", otherwise output "NO".

Sample Input

Input

7
1
2
3
8
12
24
25

Output

YES
YES
NO
YES
YES
YES
NO

Hint
这题可以把所有的情况都列举出来然后放入set，然后判断在不在set里面就行了，这里为了爆long long,有一个技巧，判断t乘上k是不是大于10^18,可以用10^18除以t，然后看结果是不是小于k，如果小于，那么t*k>10^18

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define ll __int64
#define inf 0x7fffffff
#define maxn 1000000
set<ll>myset;
set<ll>::iterator it;
ll a[maxn];
ll b[30];
void getnum()
{
int i,j;
b[1]=1;
for(i=2;i<=19;i++){
b[i]=b[i-1]*i;
}
}

void init()
{
myset.clear();
myset.insert(1);
getnum();
for(int i=2;i<=19;i++){
int tot=0;
for(it=myset.begin();it!=myset.end();it++){
tot++;
a[tot]=*it;
}
ll t;
for(int j=1;j<=tot;j++){
t=a[j];
while((1e18)/t>b[i] ){
myset.insert(t*b[i]);
t*=b[i];
}
}
}

}

int main()
{
int n,m,i,j;
init();
scanf("%d",&n);
ll num;

int cnt=0;

for(i=1;i<=n;i++){
scanf("%I64d",&num);
if(num==1){
printf("YES\n");continue;
}
it=myset.find(num);
//printf("--%d\n",*it);
if(it==myset.end())printf("NO\n");
else printf("YES\n");
}
return 0;
}