Codeforces-1249-B2. Books Exchange (hard version)（并查集/找规律）

B2. Books Exchange (hard version)

The only difference between easy and hard versions is constraints.

There are n kids, each of them is reading a unique book. At the end of any day, the i-th kid will give his book to the pi-th kid (in case of i=pi the kid will give his book to himself). It is guaranteed that all values of pi are distinct integers from 1 to n (i.e. p is a permutation). The sequence p doesn’t change from day to day, it is fixed.

For example, if n=6 and p=[4,6,1,3,5,2] then at the end of the first day the book of the 1-st kid will belong to the 4-th kid, the 2-nd kid will belong to the 6-th kid and so on. At the end of the second day the book of the 1-st kid will belong to the 3-th kid, the 2-nd kid will belong to the 2-th kid and so on.

Your task is to determine the number of the day the book of the i-th child is returned back to him for the first time for every i from 1 to n.

Consider the following example: p=[5,1,2,4,3]. The book of the 1-st kid will be passed to the following kids:

after the 1-st day it will belong to the 5-th kid,

after the 2-nd day it will belong to the 3-rd kid,

after the 3-rd day it will belong to the 2-nd kid,

after the 4-th day it will belong to the 1-st kid.

So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.

You have to answer q independent queries.

Input

The first line of the input contains one integer q (1≤q≤1000) — the number of queries. Then q queries follow.

The first line of the query contains one integer n (1≤n≤2⋅105) — the number of kids in the query. The second line of the query contains n integers p1,p2,…,pn (1≤pi≤n, all pi are distinct, i.e. p is a permutation), where pi is the kid which will get the book of the i-th kid.

It is guaranteed that ∑n≤2⋅105 (sum of n over all queries does not exceed 2⋅105).

Output

For each query, print the answer on it: n integers a1,a2,…,an, where ai is the number of the day the book of the i-th child is returned back to him for the first time in this query.

Example

input

6
5
1 2 3 4 5
3
2 3 1
6
4 6 2 1 5 3
1
1
4
3 4 1 2
5
5 1 2 4 3
           

output

1 1 1 1 1 
3 3 3 
2 3 3 2 1 3 
1 
2 2 2 2 
4 4 4 1 4 
           

解题思路：

一本书可能会经过一个循环回到某个人的手里，那么再这个循环里的所有人的天数都是一样的，应为假设是1-3-2-5-1 那么1走过3-2-5-1，那么2走过5-1-3-2

3走过2-5-1-3，天数都是相同的，所以，把他们放在同一个集合就行了。这里用了一个while把处在同一个循环里的答案变成一样的，这样每个地方最多遍历3次复杂度O（3n）。当然也可以用并查集去写。

AC代码：

/*
                 `-._:  .:'   `:::  .:\         |\__/|           /::  .:'   `:::  .:.-'
                    \      :          \         | :: |          /         :       /
                     \     ::    .     `-_______/ ::  \_______-'   .      ::   . /
                      |  :   :: ::'  :   :: ::'  ::: ::'      :: ::'  :   :: :  |
                      |     ;::         ;::        ;::         ;::         ;::  |
                      |  .:'   `:::  .:'   `:::  .:' :::  .:'   `:::  .:'   `:  |
                      /     :           :           :           :           :    \
                     /______::_____     ::    .     ::    .     ::   _____._::____\
                                    `----._:: ::'  :   :: ::'  _.----'
                                           `--.       ;::  .--'
                                               `-. .:'  .-'
                                                  \    /
                                                   \  /
                                                    \/
 */


#include <bits/stdc++.h>
//#include <iostream>
//#include <algorithm>
//#include <cstring>
//#include <cstdio>
//#include <map>
//#include <cstdlib>
#define read(x) scanf("%lld",&x)
#define re(n) for(int i = 1 ; i <= n ; i ++)
#define rev(n) for(int i = n-1 ; i >= 0 ; i --)
#define fill(x,y) memset(x,y,sizeof(x))
using namespace std;
const int N=1e6+10;
const int INF=0x3f3f3f3f;
typedef long long ll;
int a[N];
int mark[N];


int main()
{
    ll n,m,i,j,k,t = 1;
    //freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    cin>>t;
    for(int cas = 1 ; cas <= t ; cas ++)
    {
        fill(mark,0);
        cin>>n;
        re(n)  cin>>a[i];
        for(int i = 1 ; i <= n ; i ++)
        {
            if(!mark[i])
            {
                int tmp = a[i];
                int tmp1 = 1;
                while(tmp != i) {
                        tmp = a[tmp];
                        tmp1 ++;
                        //cout<<i<<" "<<tmp<<endl;
                }
                tmp = a[i];
                mark[tmp] = tmp1;
                while(tmp != i)
                {
                    tmp = a[tmp];
                    mark[tmp] = tmp1;

                }
            }
        }
        for(int i = 1 ; i < n ; i ++)
            cout<<mark[i]<<" ";
        cout<<mark[n]<<endl;
        //cout<<"Case "<<cas<<":"<<endl;
    }
    return 0;
}