//ちょっと高速解法O(nlog^2n)
#include <iostream>
#include <algorithm>
using namespace std;

int n;
int x[100000];
int dp[100000];

//返り値：x[id+1]～x[n-1]で作れる良い数列の長さのMax
int f(int id) {
	int ret = 0;
	
	if (dp[id] != -1) { return dp[id]; }
	for (int mul = 2; x[id] * mul <= 1000000; mul++) {
		int next = lower_bound(x, x + n, mul * x[id]) - x;
		if (next < n && x[next] % x[id] == 0) {
			ret = max(ret, 1 + f(next));
		}
	}
	return (dp[id] = ret);
}

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) cin >> x[i];
	sort(x, x + n);
	for (int i = 0; i < n; i++) dp[i] = -1;
	cout << 1 + f(0) << endl;
	return 0;
}