// Author: btjanaka (Bryon Tjanaka)

// Problem: (UVa) 10858

#include <bits/stdc++.h>

#define GET(x) scanf("%d", &x)

#define GED(x) scanf("%lf", &x)

typedef long long ll;

using namespace std;

// Go through factors of n, and for each factor, push it onto cur and run a new

// instance with a reduced instance of n. Analyze the children before pushing on

// the current result to keep things sorted.

void backtrack(vector<int>& cur, int n, vector<vector<int>>& res) {

for (int i = 2; i <= n; ++i) {

if (i >= cur.back() && n % i == 0) {

cur.push_back(i);

backtrack(cur, n / i, res);

cur.pop_back();

}

}

// Only push on if at least two factors (account for dummy value)

if (n == 1 && cur.size() > 2) {

res.push_back(cur);

}

}

int main() {

int n;

while (GET(n) && n) {

vector<vector<int>> res;

vector<int> cur = {INT_MIN}; // Push a small dummy value so we don't have

// to check size of cur

backtrack(cur, n, res);

printf("%d\n", (int)res.size());

for_each(res.begin(), res.end(), [](const vector<int>& factors) {

for (int i = 1; i < factors.size(); ++i) {

printf("%d%c", factors[i], i == factors.size() - 1 ? '\n' : ' ');

}

});

}

return 0;

}