Problem: You are given an integer n (n > 1).

Your task is to find a sequence of integers a_1, a_2, …, a_k such that:

  * each a_i is strictly greater than 1; 
  * a_1 ⋅ a_2 ⋅ … ⋅ a_k = n (i. e. the product of this sequence is n); 
  * a_{i + 1} is divisible by a_i for each i from 1 to k-1; 
  * k is the maximum possible (i. e. the length of this sequence is the maximum possible). 



If there are several such sequences, any of them is acceptable. It can be proven that at least one valid sequence always exists for any integer n > 1.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 5000) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (2 ≤ n ≤ 10^{10}).

It is guaranteed that the sum of n does not exceed 10^{10} (∑ n ≤ 10^{10}).

Output

For each test case, print the answer: in the first line, print one positive integer k — the maximum possible length of a. In the second line, print k integers a_1, a_2, …, a_k — the sequence of length k satisfying the conditions from the problem statement.

If there are several answers, you can print any. It can be proven that at least one valid sequence always exists for any integer n > 1.

Example

Input


4
2
360
4999999937
4998207083


Output


1
2 
3
2 2 90 
1
4999999937 
1
4998207083 

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp> //Policy Based Data Structure
// using namespace __gnu_pbds;              //Policy Based Data Structure
using namespace std;
// typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> pbds; //Policy Based Data Structure
// #define gc              getchar_unlocked
// #define pqb             priority_queue<int>
// #define pqs             priority_queue<int, vi, greater<int> >
// #define mk(arr,n,type)  type *arr = new type[n]
#define fo(i,n)         for(i=0;i<n;i++)
#define Fo(i,k,n)       for(i=k;k<n?i<n:i>n;k<n?i+=1:i-=1)
#define int             long long
#define endl            '\n'
#define w(t)            int t; cin>>t; while(t--)
#define deb(x)          cout << #x << "=" << x << endl
#define deb2(x,y)       cout << #x << "=" << x << "," << #y << "=" << y << endl
#define pb              push_back
#define mp              make_pair
#define F               first
#define S               second
#define all(x)          x.begin(), x.end()
#define clr(x)          memset(x, 0, sizeof(x))
#define sortall(x)      sort(all(x))
#define tr(it,a)        for(auto it = a.begin(); it != a.end(); it++)
#define ps(x,y)         fixed<<setprecision(y)<<x
#define setbits(x)      __builtin_popcountll(x)
#define zrobits(x)      __builtin_ctzll(x)
#define PI              3.1415926535897932384626
#define inf             1e18
// mt19937                 rng(chrono::steady_clock::now().time_since_epoch().count());  //Random Shuffler
typedef pair<int, int>   pii;
typedef vector<int>     vi;
typedef vector<pii>     vpii;
typedef vector<vi>      vvi;
typedef map<int, int>    mii;
int mpow(int base, int exp);
void ipgraph(int m);
void dfs(int u, int par);
const int mod = 1000000007;
// const int N = 3e5, M = N;
// vi g[N];
//=======================
int gcd(int a, int b)
{
	int ans = 1;
	for (int i = 2; i <= min(a, b); ++i)
	{
		if (a % i == 0 && b % i == 0)
			ans = i;
	}
	return ans;
}

int lcm(int a, int b)
{
	return a * b / gcd(a, b);
}
void sol()
{
	w(t)
	{
		int n;
		cin >> n; int prime = 1;
		for (int i = 2; i * i <= n; i++)
		{
			if (n % i == 0)
			{
				prime = 0; break;
			}
		}
		if (prime == 1)
		{
			cout << 1 << endl << n << endl;
			continue;
		}
		else
		{
			int t = n; std::map<int, int>  v;

			for (int i = 2; i * i <= n; i++)
			{
				while (t % i == 0)
				{
					t /= i;
					v[i]++;
				}
			}
			int i;
			map<int, int>::iterator itr;
			int max = 1, num = n;
			for (itr = v.begin(); itr != v.end(); ++itr)
			{
				if (itr->second > max)
				{
					max = itr->second;
					num = itr->first;
				}
			}
			if (max == 1)
			{
				cout << 1 << endl << n << endl;
				continue;

			}
			cout << max << endl; t = n;
			fo(i, max)
			{
				if (i != max - 1) {
					cout << num << " "; t /= num;
				}
				else
					cout << t << " ";
			}
			cout << endl;


			/*fo(i, v.size())
			cout << v[i] << " ";
			cout << endl;*/


		}

	}




}

int32_t main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	sol();

	return 0;
}

int mpow(int base, int exp) {
	base %= mod;
	int result = 1;
	while (exp > 0) {
		if (exp & 1) result = (result * base) % mod;
		base = (base * base) % mod;
		exp >>= 1;
	}
	return result;
}

// void ipgraph(int n, int m){
//     int i, u, v;
//     while(m--){
//         cin>>u>>v;
//         g[u-1].pb(v-1);
//         g[v-1].pb(u-1);
//     }
// }
//
// void dfs(int u, int par){
//     for(int v:g[u]){
//         if (v == par) continue;
//         dfs(v, u);
//     }
// }