You are given one integer number n. Find three distinct integers a,b,c such that 2≤a,b,c and a⋅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 t independent test cases.
Input
The first line of the input contains one integer t (1≤t≤100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2≤n≤109).
Output
For each test case, print the answer on it. Print “NO” if it is impossible to represent n as a⋅b⋅c for some distinct integers a,b,c such that 2≤a,b,c.
Otherwise, print “YES” and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823
The meaning of problems
Actually determines whether a number is decomposed into n a · b · c
Problem-solving ideas
This problem is nothing useless what algorithm, direct enumeration of violence. Note that a, b, c is less than sqrt (n * 1.0) to reduce the number of operations. Also note that the point is that digital sqrt function () must be in the floating-point type, here the card for a long time ...
Code
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int t;
int n;
cin >> t;
while (t--)
{
cin >> n;
int a = 0, b = 0;
int flag = 0;
for (int i = 2; i < sqrt(n*1.0); i++)
{
if (n % i==0)
{
if (a==0)//得出a
{
a = i;
n = n / i;
}
else//得出b,c
{
b = i;
n = n / i;
flag = 1;
break;
}
}
}
if (flag)
{
cout << "YES" << endl;
cout << a << ' ' << b << ' ' << n << endl;
}
else
cout << "NO" << endl;
}
return 0;
}
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