因为要保证 $a_{i} \vert a_{i+1}$，又要保证得到的数组最长，可以想到先把 $n$ 因数分解，得到个数最多的因子 $p$，能够保持整除关系并且最长的序列的长度就是 $p$ 的个数，其余的因子全部放在最后一个数字。

// Date: Mon Dec 11 20:15:03 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

int t;
ll n;
map<ll, int> m;

void defact(ll n) {
  for (ll i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) {
        m[i]++;
        n /= i;
      }
    }
  }
  if (n > 1)
    m[n]++;
}

int main(void) {
#ifdef _DEBUG
  freopen("1454d.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n;
    m.clear();

    defact(n);

    ll ma = 0, val = 0;
    for (auto &[x, y] : m) {
      if (y > ma) {
        ma = y;
        val = x;
      }
    }

    vector<ll> res(ma + 1, val);
    cout << ma << '\n';

    for (auto &[x, y] : m) {
      int cnt = y;
      if (x != val) {
        while (cnt--) {
          res[ma] *= x;
        }
      }
    }

    For1(i, 1, ma) { cout << res[i] << ' '; }
    cout << '\n';
  }

  return 0;
}