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:
/******************** //    Macros    // ************************/

#include "bits/stdc++.h"
using namespace std;
#define _USE_MATH_DEFINES

// #pragma GCC target("avx2")
// #pragma GCC optimization("O3")
// #pragma GCC optimization("unroll-loops")
#define endl "\n"
#define ll long long
// #define int long long
#define pl pair<long long, long long>
#define pi pair<int, int>
// #define MAX 9223372036854775807
// #define MIN -9223372036854775807
#define MAX 2147483647
#define MIN -2147483648
#define bitcount 32
#define MOD 1000000007

/******************** //    Debug    // ************************/

void DBG() {
	cerr << "]" << endl;
}
template<class H, class... T> void DBG(H h, T... t) {
	cerr << h;
	if(sizeof...(t))
		cerr << ", ";
	DBG(t...);
}
template<class H> void DBG(vector<H> v) {
	for(int i=0 ; i<(int)v.size() ; i++){
		cerr << v[i];
		if(i != (int)v.size()-1) cerr << ", ";
	}
	cerr << ']' << endl;
}
#ifdef _DEBUG
#define dbg(...) cerr << "LINE(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define dbg(...) 0
#endif

/****************** //    rand     //  *********************/

ll randNum(ll a , ll b){
	return a + rand() % (b-a+1);
}
/***************** //    template funtions    // **************/

ll mPow(ll b, ll p , ll m) {
	b %= m;
	ll result = 1;
	while(p > 0) {
		if(p & 1) result = (result * b) % m;
		b = (b * b) % m;
		p >>= 1;
	}
	return result;
}

ll gcdExtended(ll a, ll b, ll *x, ll *y) { 
    if (a == 0) { 
        *x = 0, *y = 1; 
        return b; 
    } 
    ll x1, y1;
    ll gcd = gcdExtended(b%a, a, &x1, &y1); 
    *x = y1 - (b/a) * x1; 
    *y = x1; 
    return gcd; 
} 

ll modInverse(ll b, ll m) { 
    ll x, y;
    ll g = gcdExtended(b, m, &x, &y); 
    if (g != 1) return -1; 
    return (x % m + m) % m; 
} 

ll modDivide(ll a, ll b, ll m) 
{ 
    a = a % m; 
    ll inv = modInverse(b, m); 
    if (inv == -1) 
       return -1; 
    else
       return (inv * a) % m; 
}

/** // Precompute factorial array if multiple queries are present // **/

unsigned long long nCrModPFermat(unsigned long long n, int r, int p) {
    if (n < (unsigned long long)r)
        return 0;
    if (r == 0)
        return 1;
    unsigned long long fac[n + 1];
    fac[0] = 1;
    for (unsigned long long i = 1; i <= n; i++)
        fac[i] = (fac[i - 1] * i) % p;
    return (fac[n] * modInverse(fac[r], p) % p
            * modInverse(fac[n - r], p) % p)
           % p;
}

bool sortfunc (pi a, pi b) {
	return a.first < b.first;
}

ll gcd(ll a, ll b) {
	return (b == 0) ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b) {
    return (a / gcd(a, b)) * b;
}

const int dx4[4] = {-1, 0, 1, 0}, dy4[4] = {0, -1, 0, 1};
const int dx8[8] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy8[8] = {0, 1, -1, -1, 1, 0, 1, -1};

//.................................................. 

void solve() {
	ll n;
	cin >> n;
	vector<pl>count;
	for(ll i = 2; i * i <= n; i++) {
		ll cnt = 0;
		while(n % i == 0) {
			cnt++;
			n /= i;
		}
		if(cnt > 0)
			count.push_back({cnt, i});
	}
	if(n > 1) {
		count.push_back({1, n});
	}
	sort(count.rbegin(),count.rend());
	vector<ll>ans(count[0].first,count[0].second);
	for(int i = 1; i < count.size(); i++) {
		for(int j = 0; j < count[i].first; j++) {
			ans.back() *= count[i].second;
		}
	}
	// if(n > 1) {
		
	// }
	cout << ans.size() << endl;
	for(ll i : ans) {
		cout << i << " ";
	}
	cout << endl;

}

signed main() {
// int main() {
	ios::sync_with_stdio(false);
	cout.tie(NULL);
	cin.tie(NULL);
		
	// solve();
	int t;
	cin >> t;
	while (t--) {
		solve();
	}
	return 0;
}