#include<bits/stdc++.h>
using namespace std;
using ll = __int128;

// 快速幂
ll ksm(ll a, ll b) {
    ll res = 1;
    while (b) {
        if (b & 1) res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}

// print(__int128)
void print(ll x) {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x > 9) print(x / 10);
    putchar(x % 10 + '0');
}

// 计算g(x)
ll cal(ll x) {
    if (x < 10) return 0;
    ll y = 1;
    while (x) {
        y *= x%10;
        x /= 10;
    }
    //print(y);
    //putchar('\n');
    return cal(y)+1;
}

// 通过将work(x)的各个数位相乘得到x，由x反推work(x)，不存在为-1
ll work(ll x) {
    ll res = 0;
    while (x >= 10) {
        bool ok = false;
        for (int i = 9; i >= 2; i--) {
            if (x % i == 0) {
                x /= i;
                res = res * 10 + i;
                ok = true;
                break;
            }
        }
        if (!ok) return -1;
    }
    res = res * 10 + x;
    return res;
}

void solve(){
    // 设持久度为11的数为x，f(x) = y
    // 那么y的持久度为10，且一定可以表示为2^a * 3^b * 5^c * 7^d
    for (int i = 0; i <= 63; i++) {
        ll t1 = ksm(2, i);
        for (int j = 0; ; j ++) {
            ll t2 = ksm(3, j);
            if (t1 * t2 > 1e18) break;
            for (int k = 0; ; k ++) {
                ll t3 = ksm(5, k);
                if (t1 * t2 * t3 > 1e18) break;
                for (int u = 0; ; u ++) {
                    ll t4 = ksm(7, u);
                    ll t = t1 * t2 * t3 * t4;
                    if (t > 1e18) break;
                    if (cal(t) == 10) {
                        ll t5 = work(t);
                        if (t5 != -1) {
                            print(t5);
                            putchar(' ');
                        }
                    }
                }
            }
        }
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    //int T; cin >> T;
    //while (T--)
        solve();
    return 0;
}