The score of a sequence [𝑠1,𝑠2,…,𝑠𝑑] is defined as 𝑠1⋅𝑠2⋅…⋅𝑠𝑑𝑑!, where 𝑑!=1⋅2⋅…⋅ In particular, the score of an empty sequence is 1.

For a sequence [𝑠1,𝑠2,…,𝑠𝑑], let 𝑚 be the maximum score among all its subsequences. Its cost is defined as the maximum length of a subsequence with a score of 𝑚

You are given a non-decreasing sequence [𝑎1,𝑎2,…,𝑎𝑛]of integers of length 𝑛. In other words, the condition 𝑎1≤𝑎2≤…≤𝑎𝑛is satisfied. For each 𝑘=1,2,…,𝑛, find the cost of the sequence [𝑎1,𝑎2,…,𝑎𝑘]

A sequence 𝑥 is a subsequence of a sequence 𝑦 if 𝑥 can be obtained from 𝑦 by deletion of several (possibly, zero or all) elements.

Input

Each test contains multiple test cases. The first line contains the number of test cases 𝑡(1≤𝑡≤104 The description of the test cases follows.

The first line of each test case contains an integer 𝑛 (1≤𝑛≤105) — the length of the given sequence.

The second line of each test case contains 𝑛 integers 𝑎1,𝑎2,…,𝑎𝑛 (1≤𝑎𝑖≤𝑛) — the given sequence. It is guaranteed that its elements are in non-decreasing order.

It is guaranteed that the sum of 𝑛 over all test cases does not exceed 5⋅1055⋅105.

Output

For each test case, output 𝑛� integers — the costs of sequences [𝑎1,𝑎2,…,𝑎𝑘] in ascending order of 𝑘

Example

input

Copy

3

3

1 2 3

2

1 1

5

5 5 5 5 5

output

Copy

1 1 2 
1 1 
1 2 3 4 5 

Note

In the first test case:

• The maximum score among the subsequences of [1] is 11. The subsequences [1] and [] (the empty sequence) are the only ones with this score. Thus, the cost of [1] is 1.
• The maximum score among the subsequences of [1,2] is 2 The only subsequence with this score is [2]. Thus, the cost of [1,2] is 1.
• The maximum score among the subsequences of [1,2,3] is 3. The subsequences [2,3]and [3] are the only ones with this score. Thus, the cost of [1,2,3] is 2.

Therefore, the answer to this case is 112 which are the costs of [1],[1,2] and [1,2,3] in this order.

 

判断当前序列的末尾是否小于序列长度 

#include 
#include 
#include 
#include 
#include 
using namespace std;
constexpr int N=1e5+7;
typedef long long ll;
int a[N];
int main(){int t;scanf("%d",&t);while(t--){int n;cin>>n;for(int i=1;i<=n;i++)cin>>a[i];int l=1,r=1;int len=0;while(r<=n){r++;len++;while(len>1&&a[l]