CF - Profact（DFS ＋ 剪枝）

A - Profact

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  Gym 100796K

Description

standard input/output 
Announcement

Statements

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

7
1
2
3
8
12
24
25

Output

YES
YES
NO
YES
YES
YES
NO

Hint

A factorial is any number that can be expressed as 1·2·3·...·k, for some positive integer k, and is denoted by k!.

这道题目只要简单的DFS和剪枝即可

其中剪枝主要是对于素数，众所周知，素数是无法被其他数整除的，那么如果一个数分为几个阶乘的乘积，这些阶乘里面有

素数的存在，那么一定要有素数！构成，那么如果一个素数的阶乘无法构成那么就没有必要继续递归下去了。

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define fillchar(a, x) memset(a, x, sizeof(a))
#define copy(a, b) memcpy(a, b, sizeof(a))
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r
#define FIN freopen("D://imput.txt", "r", stdin)

typedef long long LL;
typedef pair<int, int > PII;
typedef unsigned long long uLL;
template<typename T>
void print(T* p, T* q, string Gap = " ") {
    int d = p < q ? 1 : -1;
    while(p != q) {
        cout << *p;
        p += d;
        if(p != q) cout << Gap;
    }
    cout << endl;
}
template<typename T>
void print(const T &a, string bes = "") {
    int len = bes.length();
    if(len >= 2)cout << bes[0] << a << bes[1] << endl;
    else cout << a << endl;
}

const int INF = 0x3f3f3f3f;
const int MAXM = 2e5;
const int MAXN = 20 + 5;
const int mod = 1e9 + 7;
int n;
LL a;
LL X[MAXN];
int A[MAXN] = {2,3,5,7,11,13,17,19};
bool success;
void init() {
    X[0] = 1;
    for(int i = 1; i <= 20; i ++) X[i] = X[i - 1] * i;
}
void dfs(int id, LL x) {
    if(x <= 1) {
        success = true;
        return;
    }
    if(success || id <= 1) return;
    if(x % X[id] == 0) dfs(id, x / X[id]);
    if(!success) {
        for(int j = 0; j <= 7; j ++) {
            if(A[j] == id && x % A[j] == 0) return;
            if(A[j] > id) break;
        }
        dfs(id - 1, x);
    }
}
int main() {
    init();
    while(~scanf("%d", &n)) {
        for(int i = 0; i < n; i ++) {
            scanf("%I64d", &a);
            success = false;
            dfs(20, a);
            printf(success ? "YES\n":"NO\n");
        }
    }
    return 0;
}