C. Product of Three Numbers

Problem - C - Codeforces

You are given one integer number nn. Find three distinct integers a,b,ca,b,c such that 2≤a,b,c2≤a,b,c and a⋅b⋅c=na⋅b⋅c=n or say that it is impossible to do it.

If there are several answers, you can print any.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.

The next nn lines describe test cases. The ii-th test case is given on a new line as one integer nn (2≤n≤1092≤n≤109).

Output

For each test case, print the answer on it. Print "NO" if it is impossible to represent nn as a⋅b⋅ca⋅b⋅c for some distinct integers a,b,ca,b,c such that 2≤a,b,c2≤a,b,c.

Otherwise, print "YES" and any possible such representation.

Example

input

Copy

output

Copy

YES
2 4 8
NO
NO
NO
YES
3 5 823
思路：因为有三个数，如果该数的两个约数都为素数，那么不可能有三个，必定为no。若两个约束里至少有一个不是素数，那么把该数拆分成两个约数，再输出另一个数即可，但是要判断都不相等。

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
ll n,m;
int t;
bool  is_prime(ll n)//试除法判断是不是约数
{if(n==1)return false;if(n==2)return true;for(int i=2;i<=n/i;i++){if(n%i==0){return false;}}return true;
}
bool getdive(ll n,ll x)//如果不是素数，那就把这个数拆成两个约数
{for(int i=2;i<n/i;i++){if(n%i==0&&n/i!=i&&n/i!=x&&i!=x){cout<<"YES"<<endl;cout<<i<<" "<<n/i<<" ";return true;}}return false;
}
void get_divisorts(ll n)
//先找到两个约数，然后判断这两个约数一个一个判断是不是素数for(ll i=1;i<=n/i;i++){if(n%i==0&&n/i!=i){if(!is_prime(i)){if(getdive(i,n/i)){cout<<n/i<<endl;return;}}else if(!is_prime(n/i)){if(getdive(n/i,i)){cout<<i<<endl;return;}}else {cout<<"NO"<<endl;return ;}}}cout<<"NO"<<endl;return ;
}
int main()
{cin>>t;while(t--){cin>>m;get_divisorts(m);}return 0;
}

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