noname38 (lds)

#include<bits/stdc++.h>
#define ll long long
#define ld long double
#define lFOR(i, a, b) for(ll i=(a); i<=(b); i++)
#define lFORn(i, n) for(ll i=1; i<=(n); i++)
#define lFORx(i, a, b, n) for(ll i=(a); i<=(b); i+=(x))
#define FORa(i, b) for(i; i<=(b); i++)
#define FORax(i, b, x) for(i; i<=(b); i+=(x))
#define lFORR(i, a, b) for(ll i=(a); i>=(b); i--)
#define lFORRn(i, n) for(ll i=(n); i>=1; i--)
#define lFORRx(i, a, b, n) for(ll i=(a); i>=(b); i-=(x))
#define FORRa(i, b) for(i; i>=(b); i--)
#define FORRax(i, b, x) for(i; i>=(b); i-=(x))
#define iFOR(i, a, b) for(int i=(a); i<=(b); i++)
#define iFORn(i, n) for(int i=1; i<=(n); i++)
#define iFORx(i, a, b, n) for(int i=(a); i<=(b); i+=(x))
#define iFORR(i, a, b) for(int i=(a); i>=(b); i--)
#define iFORRn(i, n) for(int i=(n); i>=1; i--)
#define iFORRx(i, a, b, n) for(int i=(a); i>=(b); i-=(x))
#define pb(x) push_back(x)
#define all(x) (x).begin(), (x).end()
#define sz(x) (x).size()
#define sqr(x) (ll)(x)*(x)
#define PI (2*acos(0))
using namespace std;

template <typename T>
inline void Read(T& x)
{
    bool Neg = false;
    char c;
    for (c = getchar(); c < '0' || c > '9'; c = getchar())
        if (c == '-') Neg = !Neg;
    x = c - '0';
    for (c = getchar(); c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
    if (Neg) x = -x;
}
template <typename T>
inline void Write(T x)
{
    if (x < 0)
    {
        putchar('-'); x = -x;
    }
    T p = 1;
    for (T temp = x / 10; temp > 0; temp /= 10) p *= 10;
    for (; p > 0; x %= p, p /= 10) putchar(x / p + '0');
}






        //=================declare=================
    int n, res;
    int a[1000005], t[1000005];




//=================..end..=================
        //=================function=================





//=================..end...=================
int main(){
    freopen("LDS.inp", "r", stdin);
    freopen("LDS.out", "w", stdout);
    // ios_base::sync_with_stdio(0); cin.tie(0);
        //=================code=================
    Read(n);
    iFORn(i, n){
        cin>>a[i];
        iFORn(j, i-1) if(a[i]%a[j]==0&&t[i]<t[j]) t[i]=t[j];
        res=max(res, ++t[i]);
    }
    Write(res);





//=================end.=================
}