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 <stdio.h>
#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <list>
#include <time.h>
#include <math.h>
#include <random>
#include <deque>
#include <queue>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <chrono>
#include <cstring>
#include <bits/stdc++.h>

/* //gcc optimization
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
*/

using namespace std;

typedef long long ll;
typedef double db;

#define ft first
#define sd second
#define flag bool fl=true

#define mod 1000000000000000007
#define N 100001
#define MX 1000000001
#define MXX 1000000000000000001

#define pb push_back
#define fa(n) for(int i=0;i<n;i++)
#define fa1(n) for(int i=1;i<=n;i++)
#define fab(i, a, b) for(int i=a;i<=b;i++)
#define cs cout<<"Case "<<qq<<": "
#define csl cout<<"Case "<<qq<<":"<<endl;
#define T int tt, qq=0; cin>>tt; while(qq++<tt)
#define fast ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0)
#define ina(array, size) for(int i=0;i<size;i++) cin>>array[i];
#define outa(array, size) for(int i=0;i<size;i++) cout<<array[i]<<" ";
#define outla(array, size) for(int i=0;i<size;i++) cout<<array[i]<<endl;

#define deg(rad) (rad*180.0)/pi
#define pi acos(-1)
#define log(n) (int)ceil(log2(n))
#define logg(a)(b) log(b)/log(a)

bool fl[N+1];
vector<ll> sieve(ll n)
{
    vector<ll> primes;
    for(ll i=3; i*i<=n; i+=2)
        for(ll j=i*i; j<=n; j+=i+i)
            fl[j]=true;

    primes.pb(2);
    for(ll i=3; i<=n; i+=2) if(!fl[i]) primes.pb(i);
    return primes;
}
vector<ll> p=(sieve(N));


int main()
{
    T
    {
        ll n, nn;
        cin>>n;
        nn=n;

        int mx=0, mxp=-1;
        for(int i=0;i<p.size();i++)
        {
            if(n%p[i]==0)
            {
                int cnt=0;
                while(n%p[i]==0) n/=p[i], cnt++;
                if(cnt>mx) mx=cnt, mxp=p[i];
            }
        }

        if(mxp==-1) cout<<1<<endl<<n<<endl;
        else
        {
            cout<<mx<<endl;
            for(int i=1;i<mx;i++)
            {
                cout<<mxp<<" ";
                nn/=mxp;
            }
            cout<<nn<<endl;
        }
    }
}

/*
//----------useful_functions----------

long long extended_euclid(ll a, ll b, ll &x, ll &y) //ax+by=c
{
    if(b==0)
    {
        x=1, y=0;
        return a;
    }

    ll x1, y1;
    ll d=extended_euclid(b, a%b, x1, y1);
    x=y1;
    y=x1-y1*(a/b);
    return d;
}

long long bigMod(ll a, ll b) //(a^b)%mod
{
    ll ans=1;
    while(b)
    {
        if(b%2==1) ans=(ans*a)%mod;
        a=(a*a)%mod;
        b/=2;
    }
    return ans%mod;
}

long long modInverse(ll a, ll b) //a*x=1 mod b
{
    ll x, y;
    ll gcd=extended_euclid(a, b, x, y);

    if(gcd==1) return ((x%b)+b)%b;
    else return 1;
}

long long nCr(ll n, ll r) //n<=1e6
{
    //declare globally-> ll fact[1000001];
    fact[1]=1;
    for(ll i=2; i<1000001; i++)
        fact[i]=(i*fact[i-1])%mod;

    return (fact[n]*modInverse(fact[r], mod)*modInverse(fact[n-r], mod))%mod;
}

long long josephus(ll n, ll k) //k'th person is killed out of n in the circle
{
    if(n==1) return 1;
    else return (josephus(n-1, k)+k-1)%n+1;
}

bool fl[N+1];
vector<ll> sieve(ll n)
{
    vector<ll> primes;
    for(ll i=3; i*i<=n; i+=2)
        for(ll j=i*i; j<=n; j+=i+i)
            fl[j]=true;

    primes.pb(2);
    for(ll i=3; i<=n; i+=2) if(!fl[i]) primes.pb(i);
    return primes;
}
vector<ll> p=(sieve(N));

ll eulerTotient(ll x)
{
    db ans=x;
    for(int i=0; i<p.size() && p[i]*p[i]<=x; i++)
    {
        if(x%p[i]==0)
        {
            ans*=(1.0-(1/(db)p[i]));
            while(x%p[i]==0) x/=p[i];
        }
    }
    if(x>1) ans*=(1.0-(1/(db)x));

    return ans;
}

//<---query sort--->

//global part
const ll block=1300;
struct query
{
    ll id, l, r;
}qq[N];

//main function part
    cin>>q;---->total q queries
    for(int i=1;i<=q;i++)
    {
        cin>>qq[i].l>>qq[i].r;
        qq[i].id=i;
    }
    sort(qq+1, qq+1+q, [](query a, query b){ return a.l/block==b.l/block ? a.r<b.r : a.l/block<b.l/block;});

//---->end<----

*/