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.*;
    import java.util.*;
    import java.math.*;
     
    public class bhaa {
     
        InputStream is;
        PrintWriter o;
        /////////////////// CODED++ BY++ ++ ++ ++ BHAVYA++ ARORA++ ++ ++ ++ FROM++ JAYPEE++ INSTITUTE++ OF++ INFORMATION++ TECHNOLOGY++ ////////////////
     
    ///////////////////////// Make it work, make it right, make it fast. Make it work, make it right, make it fast. Make it work, make it right, make it fast. Make it work, make it right, make it fast. /////////////////
     
    
        
        void solve() {
        	int te=ni();
        	while(te-->0)
        	{
        		int n=ni();
        		int arr[]=nia(n);
        		int res[]=new int[n];
        		for(int i=0;i<n;i++)
        		{
        			if(res[i]==0)
        			{
        				if((arr[i]!=i+1))
        				{
        					int t=i;
        					int d=1;
        					while(arr[t]!=i+1)
        					{
        						t=arr[t]-1;
        						d++;

        					}
        					res[i]=d;
        					t=i;
        					while(arr[t]!=i+1)
        					{
        						t=arr[t]-1;
        						res[t]=d;
        					}


        				}
        				else
        				{
        					res[i]=1;
        				}
        			}
        		}
        		for(int i=0;i<n;i++)
        		{
        			o.print(res[i]+" ");

        		}
        		o.println();


        	}

     
        
     
            
            
     
            
     
     
        }
    
     
     
     
     
     
     
       
     
     
        
     
     
     
     
     
     
     
     
     
     
     
     
        
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
        //---------- I/O Template ----------
        
        public static void main(String[] args) { new bhaa().run(); }
        void run() { 
            is = System.in; 
            o = new PrintWriter(System.out);
            solve();
            o.flush();
        }
        
        byte input[] = new byte[1024];
        int len = 0, ptr = 0;
        
        int readByte() { 
            if(ptr >= len) { ptr = 0; 
                try { len = is.read(input); } 
                catch(IOException e) { throw new InputMismatchException(); } 
                if(len <= 0) { return -1; } 
            } return input[ptr++];
        }
        boolean isSpaceChar(int c) { return !( c >= 33 && c <= 126 ); }
        int skip() { 
            int b = readByte(); 
            while(b != -1 && isSpaceChar(b)) { b = readByte(); } 
            return b;
        }
        
        char nc() { return (char)skip(); }
        String ns() { 
            int b = skip(); 
            StringBuilder sb = new StringBuilder(); 
            while(!isSpaceChar(b)) { sb.appendCodePoint(b); b = readByte(); } 
            return sb.toString();
        }
        String nLine() { 
            int b = skip(); 
            StringBuilder sb = new StringBuilder(); 
            while( !(isSpaceChar(b) && b != ' ') ) { sb.appendCodePoint(b); b = readByte(); } 
            return sb.toString();
        }
        int ni() { 
            int n = 0, b = readByte(); 
            boolean minus = false; 
            while(b != -1 && !( (b >= '0' && b <= '9') || b == '-')) { b = readByte(); } 
            if(b == '-') { minus = true; b = readByte(); } 
            if(b == -1) { return -1; }  //no input 
            while(b >= '0' && b <= '9') { n = n * 10 + (b - '0'); b = readByte(); } 
            return minus ? -n : n;
        }
        long nl() { 
            long n = 0L;    int b = readByte(); 
            boolean minus = false; 
            while(b != -1 && !( (b >= '0' && b <= '9') || b == '-')) { b = readByte(); } 
            if(b == '-') { minus = true; b = readByte(); } 
            while(b >= '0' && b <= '9') { n = n * 10 + (b - '0'); b = readByte(); } 
            return minus ? -n : n;
        }
     
        double nd() { return Double.parseDouble(ns()); }
        float nf() { return Float.parseFloat(ns()); }
        int[] nia(int n) { 
            int a[] = new int[n]; 
            for(int i = 0; i < n; i++) { a[i] = ni(); } 
            return a;
        }
        long[] nla(int n) { 
            long a[] = new long[n]; 
            for(int i = 0; i < n; i++) { a[i] = nl(); } 
            return a;
        }
        int [][] nim(int n)
        {
            int mat[][]=new int[n][n];
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<n;j++)
                {
                    mat[i][j]=ni();
                }
            }
            return mat;
        }
        long [][] nlm(int n)
        {
            long mat[][]=new long[n][n];
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<n;j++)
                {
                    mat[i][j]=nl();
                }
            }
            return mat;
        }
        
     
     
     
     
        
        char[] ns(int n) { 
            char c[] = new char[n]; 
            int i, b = skip(); 
            for(i = 0; i < n; i++) { 
                if(isSpaceChar(b)) { break; } 
                c[i] = (char)b; b = readByte(); 
            } return i == n ? c : Arrays.copyOf(c,i);
        }
        void piarr(int arr[])
        {
            for(int i=0;i<arr.length;i++)
            {
                o.print(arr[i]+" ");
     
            }
            o.println();
        }
        void plarr(long arr[])
        {
            for(int i=0;i<arr.length;i++)
            {
                o.print(arr[i]+" ");
     
            }
            o.println();
        }
        
        void pimat(int mat[][])
        {
            for(int i=0;i<mat.length;i++)
            {
                for(int j=0;j<mat[0].length;j++)
                {
                    o.print(mat[i][j]);
                }
                o.println();
            }
        }
        void plmat(long mat[][])
        {
            for(int i=0;i<mat.length;i++)
            {
                for(int j=0;j<mat[0].length;j++)
                {
                    o.print(mat[i][j]);
                }
                o.println();
            }
     
        }
     
     
     
     
        //////////////////////////////////// template finished //////////////////////////////////////
     
     
     
    }