B. Orac and Models

There are nn models in the shop numbered from 11 to nn, with sizes s1,s2,…,sns1,s2,…,sn.

Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).

Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices ijij and ij+1ij+1 (note that ij<ij+1ij<ij+1, because Orac arranged them properly), ij+1ij+1 is divisible by ijij and sij<sij+1sij<sij+1.

For example, for 66 models with sizes {3,6,7,7,7,7}{3,6,7,7,7,7}, he can buy models with indices 11, 22, and 66, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.

Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.

Input

The first line contains one integer t (1≤t≤100)t (1≤t≤100): the number of queries.

Each query contains two lines. The first line contains one integer n (1≤n≤100000)n (1≤n≤100000): the number of models in the shop, and the second line contains nn integers s1,…,sn (1≤si≤109)s1,…,sn (1≤si≤109): the sizes of models.

It is guaranteed that the total sum of nn is at most 100000100000.

Output

Print tt lines, the ii-th of them should contain the maximum number of models that Orac can buy for the ii-th query.

Example

input

Copy

4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9

output

Copy

2
3
1
1

Note

In the first query, for example, Orac can buy models with indices 22 and 44, the arrangement will be beautiful because 44 is divisible by 22 and 66 is more than 33. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.

In the second query, Orac can buy models with indices 11, 33, and 66. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.

In the third query, there are no beautiful arrangements with more than one model.

从前往后更新，复杂度O(nlogn)

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define pb              push_back
#define mk              make_pair
using namespace std;
int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const int mod = 998244353;
const int N = 2e5 + 5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}

ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m * n / gcd(m, n);
}

int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        int n;
        cin >> n;
        vector<int> a(n+1),b(n+1,1);
        rep(i, 1, n)
        {
            cin >> a[i];
        }
        rep(i, 1, n)
        {
            for(int j=2*i;j<=n;j+=i)
            {
                if (a[j] > a[i])
                {
                    b[j] = max(b[j], b[i] + 1);
                }
            }
        }
        int ans = 0;
        rep(i, 1, n)
        {
            ans = max(ans, b[i]);
        }
        cout << ans << endl;
    }
    return 0;
}