Discription

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,?...,?a_n}as follows. The vertices of the given graph are numbers from set A, and two numbers a_i and a_j (i?≠?j) are connected by an edge if and only if either a_i is divisible by a_j, or a_j is divisible by a_i.

You are given a set of non-negative integers 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,?...,?a_n (1?≤?a_i?≤?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

8
3 4 6 8 10 18 21 24

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.

我們不妨把邊看成有向的，當 a[i]|a[j] 時從 i 向 j 連邊（要先把重復元素縮成一個點），這樣我們發現隨便一條鏈上的元素 都可以成為一個團，並且這是一個DAG。

所以我們直接找最長鏈好了。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1000000;
int n,cnt[maxn+5],f[maxn+5];
inline int read(){
	int x=0; char ch=getchar();
	for(;!isdigit(ch);ch=getchar());
	for(;isdigit(ch);ch=getchar()) x=x*10+ch-‘0‘;
	return x;
}
int main(){
	n=read();
	for(int i=1;i<=n;i++) cnt[read()]++;
	for(int i=maxn;i;i--){
		for(int j=i;j<=maxn;j+=i) f[i]=max(f[i],f[j]);
		f[i]+=cnt[i];
	}
	printf("%d\n",f[1]);
	return 0;
}

CodeForces - 566F Clique in the Divisibility Graph