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 a_i, the cost on the i-th bill (1 ≤ a_i ≤ 10¹⁸).

Output

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

Sample Input

Input

71238122425

Output

YESYESNOYESYESYESNO

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 1000000set<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;}