The only difference between easy and hard versions is constraints.

There are nn kids, each of them is reading a unique book. At the end of any day, the ii-th kid will give his book to the pipi-th kid (in case of i=pii=pi the kid will give his book to himself). It is guaranteed that all values of pipi are distinct integers from 11 to nn (i.e. pp is a permutation). The sequence pp doesn't change from day to day, it is fixed.

For example, if n=6n=6 and p=[4,6,1,3,5,2]p=[4,6,1,3,5,2] then at the end of the first day the book of the 11-st kid will belong to the 44-th kid, the 22-nd kid will belong to the 66-th kid and so on. At the end of the second day the book of the 11-st kid will belong to the 33-th kid, the 22-nd kid will belong to the 22-th kid and so on.

Your task is to determine the number of the day the book of the ii-th child is returned back to him for the first time for every ii from 11 to nn.

Consider the following example: p=[5,1,2,4,3]p=[5,1,2,4,3]. The book of the 11-st kid will be passed to the following kids:

  • after the 11-st day it will belong to the 55-th kid,
  • after the 22-nd day it will belong to the 33-rd kid,
  • after the 33-rd day it will belong to the 22-nd kid,
  • after the 44-th day it will belong to the 11-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 qq independent queries.

Input

The first line of the input contains one integer qq (1≤q≤10001≤q≤1000) — the number of queries. Then qq queries follow.

The first line of the query contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of kids in the query. The second line of the query contains nn integers p1,p2,…,pnp1,p2,…,pn (1≤pi≤n1≤pi≤n, all pipi are distinct, i.e. pp is a permutation), where pipi is the kid which will get the book of the ii-th kid.

It is guaranteed that ∑n≤2⋅105∑n≤2⋅105 (sum of nn over all queries does not exceed 2⋅1052⋅105).

Output

For each query, print the answer on it: nn integers a1,a2,…,ana1,a2,…,an, where aiai is the number of the day the book of the ii-th child is returned back to him for the first time in this query.

Example
input

Copy
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

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


#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 <unordered_set>
#include <unordered_map>
#define ll long long
using namespace std;
int dir[][] = { {,},{,-},{-,},{,} };
vector<int> p,help,res;
int cnt;
void dfs(int start, int pos)
{
help.push_back(pos);
cnt++;
int t = p[pos];
p[pos] = ;
if (start == t)
return;
dfs(start, t);
} int main()
{
int q;
cin >> q;
while (q--)
{
int n;
cin >> n;
p.resize(n+);
res.resize(n + ); for (int i = ; i <= n; i++)
cin >> p[i];
for (int i = ; i <= n; i++)
{
help.clear();
cnt = ;
if (p[i] == )
continue;
dfs(i, i);
for (int j = ; j < help.size(); j++)
{
res[help[j]] = cnt;
}
}
for (int i = ; i < res.size(); i++)
cout << res[i] << " ";
cout << endl;
}
//system("pause");
return ;
}

最新文章

  1. ARM概论(Advanced RISC Machines)
  2. ppmoney
  3. [Java Web]Error parsing HTTP request header Note: further occurrences of HTTP header parsing errors
  4. Django数据库设置
  5. 12-27cell常用的属性
  6. hexo系列教程
  7. 【原】linux系统运维工具必备
  8. 从头开始建网站(三)DNS
  9. BHuman文档结构
  10. Python基础之字符编码
  11. Sql语句varchar或nvarchar字段条件前加N的性能差异
  12. Vue-route实现原理
  13. SpringBoot入门教程(七)整合themeleaf+bootstrap
  14. python-day7-静态方法、类方法、属性方法、特殊成员方法、反射、异常处理、socket
  15. goland 文件头注释
  16. [2016北京集训测试赛17]crash的游戏-[组合数+斯特林数+拉格朗日插值]
  17. Spring JDBC更新数据
  18. shell逻辑运算符
  19. Beand的高级特征
  20. Monit安装与配置

热门文章

  1. 手动部署:在eclipse导入web项目并更新包到本地部署
  2. Parity game POJ - 1733 带权并查集
  3. Firefox下载.net服务器文件时中文乱码
  4. openssl 生成免费证书
  5. Dev 控件笔记1 repositoryItemLookUpEdit 控件
  6. CentOS 7在执行yum操作时 报错
  7. fastadmin选择下拉框
  8. openlayers 保存当前地图View为图片
  9. Attention machenism
  10. arm学习笔记