Problem: 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 p_i-th kid (in case of i = p_i the kid will give his book to himself). It is guaranteed that all values of p_i 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 ⋅ 10^5) — the number of kids in the query. The second line of the query contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i are distinct, i.e. p is a permutation), where p_i is the kid which will get the book of the i-th kid.

It is guaranteed that ∑ n ≤ 2 ⋅ 10^5 (sum of n over all queries does not exceed 2 ⋅ 10^5).

Output

For each query, print the answer on it: n integers a_1, a_2, ..., a_n, where a_i 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 

Solution (in JAVA):

Solution:

import java.util.*;




import java.io.*;
public class Solution {
	static FastScanner scr=new FastScanner();
	static PrintStream out=new PrintStream(System.out);
	static StringBuilder sb=new StringBuilder();
	
	static long gcd(long a,long b){
		if(b==0) {
			return a;
		}
			
		return gcd(b,a%b);
	}
	
	
	static class pair{
		int x;int y;
		pair(int x,int y){ 
			this.x=x;this.y=y;
			
		}
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
	 	String next() {
	 		while (!st.hasMoreTokens())
	 			try {
	 				st=new StringTokenizer(br.readLine());
	 			} catch (IOException e) {
	 				e.printStackTrace();
	 			}
	 			return st.nextToken();
		}
	 	int[]  sort(int a[]) {
			 
	        int multiplier = 1, len = a.length, max = Integer.MIN_VALUE;
	        int b[] = new int[len];
	        int bucket[];
	        for (int i = 0; i < len; i++) if (max < a[i]) max = a[i];
	        while (max / multiplier > 0) {
	            bucket = new int[10];
	            for (int i = 0; i < len; i++) bucket[(a[i] / multiplier) % 10]++;
	            for (int i = 1; i < 10; i++) bucket[i] += (bucket[i - 1]);
	            for (int i = len - 1; i >= 0; i--) b[--bucket[(a[i] / multiplier) % 10]] = a[i];
	            for (int i = 0; i < len; i++) a[i] = b[i];
	            multiplier *= 10;
	        }
	        return a;
	    }
	 	 long modPow(long base,long exp,long mod) {
			if(exp==0) {
				return 1;
			}
			if(exp%2==0) {
				long res=(modPow(base,exp/2,mod));
				return (res*res)%mod;
			}
			return (base*modPow(base,exp-1,mod))%mod;
		}
		int []reverse(int a[]){
			int b[]=new int[a.length];
			int index=0;
			for(int i=a.length-1;i>=0;i--) {
				b[index]=a[i];
			}
			return b;
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long[] readLongArray(int n) {
			long  [] a=new long  [n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	
	static void print(Object ele) {
		sb.append(ele).append(" ");
	}
	static void println(Object ele) {
//		AccessBridge.
		sb.append(ele).append('\n');
		
	}
	static void println() {
		sb.append('\n');
	}
	static long LCM(long a,long b) {
		return a*b/(gcd(a, b));
	}
//	static long n,m,k,l,q,count,sum,temp,ans;
//	static long []a;
//	static long b[];
//	static  char c[];
//	static char t[]
	static void solve() {
		int n=scr.nextInt();
		int []a=scr.readArray(n);
		
//		ArrayList<Integer>index[]=new ArrayList[n+1];
		
		int count[]=new int[n+1];
		boolean visited[]=new boolean[n+1];
		
		for(int i=0;i<n;i++) {
//			int cnt=0;
			if(visited[i]==true) {
				continue;
			}
			int ele=a[i];
			ArrayList<Integer>list=new ArrayList<>();
//			list
			while(ele-1!=i) {
				list.add(ele);
				ele=a[ele-1];
				visited[ele-1]=true;
			}
			for(Integer k:list) {
				count[k]=list.size();
			}
		}
		
		for(int i=1;i<=n;i++) {
//			out.print((count[i]+1)+" ");
			sb.append(count[i]+1).append(" ");
		}
		sb.append('\n');
	}
	
	static int  MAX=Integer.MAX_VALUE;
	static int  MIN=Integer.MIN_VALUE;
	public static void main(String []args) {
		int t=scr.nextInt();
		while(t-->0) {
			solve();
		}
//		solve();
		out.println(sb);
	}
}