​​http://codeforces.com/contest/566/problem/F​​

F. Clique in the Divisibility Graph

time limit per test

memory limit per test

input

output

As you must know, the maximum clique problem in an arbitrary graph is NP-hard. Nevertheless, for some graphs of specific kinds it can be solved effectively.

Just in case, let us remind you that a clique in a non-directed graph is a subset of the vertices of a graph, such that any two vertices of this subset are connected by an edge. In particular, an empty set of vertexes and a set consisting of a single vertex, are cliques.

Let's define a divisibility graph for a set of positive integers A = {a1, a2, ..., an} as follows. The vertices of the given graph are numbers from set A, and two numbers ai and aj (i ≠ j) are connected by an edge if and only if either ai is divisible by aj, or aj is divisible by ai.

You are given a set of non-negative integers A. Determine the size of a maximum clique in a divisibility graph for set A.

Input

The first line contains integer n (1 ≤ n ≤ 106), that sets the size of set A.

The second line contains n distinct positive integers a1, a2, ..., an (1 ≤ ai ≤ 106) — elements of subset A. The numbers in the line follow in the ascending order.

Output

Print a single number — the maximum size of a clique in a divisibility graph for set A.

Examples

input

8
3 4 6 8 10 18 21 24

output

3

Note

In the first sample test a clique of size 3 is, for example, a subset of vertexes {3, 6, 18}. A clique of a larger size doesn't exist in this graph.

 

设DP[i]表示第i个数字结尾的最大合法情况。

那么递推过来的话，要在前i - 1个数字中，是a[i]约数的，才能递推过来dp[i]，那么需要把a[i]分解因子。这样要sqrtn复杂度。超时。

可以考虑以第a[i]个数递推去后面，就是去更新[i + 1，以后的数字。

那么只有k * a[i]的才能递推过去。

但是有些会重复，

3 4 24这样，24既可以从3递推过来，也可以从4递推过来。那么娶个max即可。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 1e6 + 20;
int a[maxn];
int add[maxn];
int calc(int val) {
    int en = (int)sqrt(val * 1.0);
    int res = 0;
    for (int i = 2; i <= en; ++i) {
        if (val % i == 0) {
            res = max(res, add[i]);
            res = max(res, add[val / i]);
        }
    }
    return res;
}
void work() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
        add[a[i]] += 1;
        for (int j = 2 * a[i]; j <= maxn - 20; j += a[i]) {
            add[j] = max(add[j], add[a[i]]);
        }
        ans = max(ans, add[a[i]]);
    }
//    for (int i = 1; i <= n; ++i) {
//        printf("%d ", add[a[i]]);
//    }
    printf("%d\n", ans);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return 0;
}

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