D. Number into Sequence

// Problem: D. Number into Sequence
// Contest: Codeforces - Codeforces Round #686 (Div. 3)
// URL: https://codeforces.com/problemset/problem/1454/D
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// 2022-02-22 16:55:51
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}

bool check (ll tar) {
    if (tar < 2) return 0;
    for (int i = 2; i <= tar / i; i ++)
        if (tar % i == 0) return 0;
    return 1;
}
vector<ll> ans, ans2;
void solve() {
    ans.clear();
    ans2.clear();
    ll n;
    cin >> n;
    if (check(n)) {
        cout << 1 << endl;
        cout << n << endl;
    }
    else {
        ll tn = n;
        for (int i = 2; i <= n/ i; i ++)
            if (n % i == 0) {
                while (n % i == 0) {
                    ans.pb(i);
                    n /= i;
                }
            }
        if (n > 1) ans.pb(n);
        map<ll, ll> c;
        for (auto t : ans) c[t] ++;
        ll temp = 1;
        int id = -1;
        int cnt = 0;
         for (int i = 0; i < ans.size() - 1; i ++)
            if (ans[i + 1] % ans[i] == 0) {
                if (c[ans[i]] > cnt) {
                    cnt = c[ans[i]];
                    id = i;
                }
            }
        
        if (id == -1) {
            cout << 1 <<  endl;
            cout << tn << endl;
            return;
        }
        // for (auto t : ans) cout << t << " ";
        // puts("");
        for (int i = id; i < ans.size() - 1; i ++) {
            if (ans[i + 1] % ans[i] == 0) {
                ans2.pb(ans[i]);
                
                if (i == ans.size() - 2) {
                    temp *= ans[i];
                    ans2.pb(tn / temp);
                    break;
                }
            }
            else {
                ans2.pb(tn / temp);             
                break;
            }
            temp *= ans[i];
        }
        cout << ans2.size() << endl;
        for (auto t : ans2) cout << t << " ";
        puts("");
    }
}
int main () {
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}

优先选择一个出现次数最多的因子，作为开头。