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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedHashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.StringTokenizer;
 
 
public class hello1
 {  
	static class FastReader 
       { 
           BufferedReader br; 
           StringTokenizer st; 
     
           public FastReader() 
           { 
               br = new BufferedReader(new
                        InputStreamReader(System.in)); 
           } 
     
           String next() 
           { 
               while (st == null || !st.hasMoreElements()) 
               { 
                   try
                   { 
                       st = new StringTokenizer(br.readLine()); 
                   } 
                   catch (IOException  e) 
                   { 
                       e.printStackTrace(); 
                   } 
               } 
               return st.nextToken(); 
           } 
     
           int nextInt() 
           { 
               return Integer.parseInt(next()); 
           } 
     
           long nextLong() 
           { 
               return Long.parseLong(next()); 
           } 
     
           double nextDouble() 
           { 
               return Double.parseDouble(next()); 
           } 
     
           String nextLine() 
           { 
               String str = ""; 
               try
               { 
                   str = br.readLine(); 
               } 
               catch (IOException e) 
               { 
                   e.printStackTrace(); 
               } 
               return str; 
           } 
         
       }
    
    static String sum (String s)
    {
        String s1 = ""; 
        
        if(s.contains("a"))
            s1+="a";
        if(s.contains("e"))
            s1+="e";
        if(s.contains("i"))
            s1+="i";
        if(s.contains("o"))
            s1+="o";
        if(s.contains("u"))
            s1+="u";
        
            return s1;
            
        
    }
    
 
    public static HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm) 
    { 
        // Create a list from elements of HashMap 
        List<Map.Entry<Integer, Integer> > list = 
               new LinkedList<Map.Entry<Integer, Integer> >(hm.entrySet()); 
  
        // Sort the list 
        Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() { 
            public int compare(Map.Entry<Integer, Integer> o1,  
                               Map.Entry<Integer, Integer> o2) 
            { 
                return (o1.getValue()).compareTo(o2.getValue()); 
            } 
        }); 
          
        // put data from sorted list to hashmap  
        HashMap<Integer, Integer> temp = new LinkedHashMap<Integer, Integer>(); 
        for (Map.Entry<Integer, Integer> aa : list) { 
            temp.put(aa.getKey(), aa.getValue()); 
        } 
        return temp; 
    } 
    
 
public static void main(String args[])
    {
        FastReader input=new FastReader();
       
        int q = input.nextInt();
        
        while(q-->0)
        {
        	int n = input.nextInt();
        	int a[] = new int[n];
        	for(int i=0;i<n;i++)
        	{
        		a[i] = input.nextInt();
        	}
        	int count=0;
        	int x = 1;
        	
        	int b[] = new int[n];
        	
        	int sum[] = new int[n];
        	
        	
        	for(int j=0;j<n;j++)
        	{
        		 count=1;
        		 x =j+1;
        		if(b[j]==0)
        		{
        			b[j] = j+1;
        			for(int i=0;i<n;i++)
        			{
        				if(j+1 == a[x-1])
        				{
        					
        					sum[j] = count;
        					break;
        				}
        				else
        				{
        					b[x-1] = j+1;
        					x = a[x-1];
        					count++;
        				}
        		
        			}	
        		}
        	
        
        }
        	for(int i=0;i<n;i++)
        	{
        		System.out.print(sum[b[i]-1] + " ");
        	}
        	System.out.println();
        }
        
        
    }
        
        
static int modFact(int n, 
        int p) 
{ 
if (n >= p) 
return 0; 

int result = 1; 
for (int i = 1; i <= n; i++) 
result = (result * i) % p; 

return result; 
}




static int binarySearch(long arr[], long x) 
{ 
    int l = 0, r = arr.length - 1; 
    while (l <= r) { 
        int m = l + (r - l) / 2; 

        // Check if x is present at mid 
        if (arr[m] == x) 
            return m; 

        // If x greater, ignore left half 
        if (arr[m] < x) 
            l = m + 1; 

        // If x is smaller, ignore right half 
        else
            r = m - 1; 
    } 

    // if we reach here, then element was 
    // not present 
    return -1; 
}
    

static long gcd(long a, long b) 
{ 
  if (b == 0) 
    return a; 
  return gcd(b, a % b);  
} 


public static void sort(int arr[])
{
    int n = arr.length;
    int count =0;
    // One by one move boundary of unsorted subarray
    for (int i = 0; i < n-1; i++)
    {
        // Find the minimum element in unsorted array
        int min_idx = i;
        for (int j = i+1; j < n; j++)
            if (arr[j] < arr[min_idx])
                min_idx = j;
 
        // Swap the found minimum element with the first
        // element
        if(min_idx!=i)
        {
        	int temp = arr[min_idx];
        
        arr[min_idx] = arr[i];
        arr[i] = temp;
        count++;
        }
        }
    System.out.println(count);
}


   
public static int even_element(int b[],int n)
{
	int c[] = new int[n/2 + 1];
	int l=0;
	for(int i=1;i<n;i+=2)
	{
		c[l] = b[i];
		
		System.out.print(c[l] + " ");
		l++;
	}
	System.out.println();
	if(c.length!=1)
	{
		return even_element(c,n/2);
	}
	else
		return c[0];
}
 
  
    
    static boolean isPrime(int n) 
    { 
        // Corner case 
        if (n <= 1) 
            return false; 
  
        // Check from 2 to n-1 
        for (int i = 2; i < n; i++) 
            if (n % i == 0) 
                return false; 
  
        return true; 
    } 
 }