Problem: Example

Input

20


Output

4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto& (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define sar(a,n) cout<<#a<<":";rep(kbrni,n)cout<<" "<<a[kbrni];cout<<endl
#define svec(v) cout<<#v<<":";rep(kbrni,v.size())cout<<" "<<v[kbrni];cout<<endl
#define svecp(v) cout<<#v<<":";each(kbrni,v)cout<<" {"<<kbrni.first<<":"<<kbrni.second<<"}";cout<<endl
#define sset(s) cout<<#s<<":";each(kbrni,s)cout<<" "<<kbrni;cout<<endl
#define smap(m) cout<<#m<<":";each(kbrni,m)cout<<" {"<<kbrni.first<<":"<<kbrni.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 100005;

map<ll,ll> mp[15][130];
map<ll,ll> mp2[15][130][3];

void solve(int& sm,ll& mul,string& s)
{
    sm = 0,mul = 1;
    rep(i,len(s)){
        sm += (s[i]-'0');
        mul *= (s[i]-'0')+1;
    }
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    string s;
    cin >> s;
    int keta = len(s);
    int sm;
    ll mul;
    solve(sm,mul,s);
    mp2[0][0][0][1] = 1;
    rep(i,keta){
        rep(j,sm+1){
            each(it,mp2[i][j][0]){
                rep(k,min(sm+1-j,10)){
                    if(k < s[i]-'0'){
                        mp2[i+1][j+k][1][it.fi*(k+1)] += it.se;
                    }else if(k == s[i]-'0'){
                        mp2[i+1][j+k][0][it.fi*(k+1)] += it.se;
                    }else{
                        mp2[i+1][j+k][2][it.fi*(k+1)] += it.se;
                    }
                }
            }
            srep(k,1,3){
                each(it,mp2[i][j][k]){
                    rep(l,min(sm+1-j,10)){
                        mp2[i+1][j+l][k][it.fi*(l+1)] += it.se;
                    }
                }
            }
        }
    }
    ll ans = 0;
    rep(i,sm){
        rep(j,3){
            each(it,mp2[keta][i][j]){
                ans += it.se;
            }
        }
    }
    rep(i,3){
        each(it,mp2[keta][sm][i]){
            if(it.fi >= mul){
                break;
            }
            ans += it.se;
        }
    }
    ans += mp2[keta][sm][1][mul];
    cout << ans << "\n";
    return 0;
}