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:
//package credit;


import java.awt.*;
import java.beans.IntrospectionException;
import java.io.PrintWriter;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.sql.SQLIntegrityConstraintViolationException;
import java.util.*;
import java.util.List;

public class sas {
    static boolean v[];
    static int ans[];
    int size[];
    static int count=0;
    static int dsu=0;
    static int c=0;
    int min=Integer.MAX_VALUE;
    int max=Integer.MIN_VALUE;
    boolean aBoolean=true;
    PrintWriter printWriter = new PrintWriter(System.out);
    static class Reader
        {
            final private int BUFFER_SIZE = 1 << 16;
            private DataInputStream din;
            private byte[] buffer;
            private int bufferPointer, bytesRead;

            public Reader()
            {
                din = new DataInputStream(System.in);
                buffer = new byte[BUFFER_SIZE];
                bufferPointer = bytesRead = 0;
            }

            public Reader(String file_name) throws IOException
            {
                din = new DataInputStream(new FileInputStream(file_name));
                buffer = new byte[BUFFER_SIZE];
                bufferPointer = bytesRead = 0;
            }

            public String readLine() throws IOException
            {
                byte[] buf = new byte[64]; // line length
                int cnt = 0, c;
                while ((c = read()) != -1)
                {
                    if (c == '\n')
                        break;
                    buf[cnt++] = (byte) c;
                }
                return new String(buf, 0, cnt);
            }

            public int nextInt() throws IOException
            {
                int ret = 0;
                byte c = read();
                while (c <= ' ')
                    c = read();
                boolean neg = (c == '-');
                if (neg)
                    c = read();
                do
                {
                    ret = ret * 10 + c - '0';
                }  while ((c = read()) >= '0' && c <= '9');

                if (neg)
                    return -ret;
                return ret;
            }

            public long nextLong() throws IOException
            {
                long ret = 0;
                byte c = read();
                while (c <= ' ')
                    c = read();
                boolean neg = (c == '-');
                if (neg)
                    c = read();
                do {
                    ret = ret * 10 + c - '0';
                }
                while ((c = read()) >= '0' && c <= '9');
                if (neg)
                    return -ret;
                return ret;
            }

            public double nextDouble() throws IOException
            {
                double ret = 0, div = 1;
                byte c = read();
                while (c <= ' ')
                    c = read();
                boolean neg = (c == '-');
                if (neg)
                    c = read();

                do {
                    ret = ret * 10 + c - '0';
                }
                while ((c = read()) >= '0' && c <= '9');

                if (c == '.')
                {
                    while ((c = read()) >= '0' && c <= '9')
                    {
                        ret += (c - '0') / (div *= 10);
                    }
                }

                if (neg)
                    return -ret;
                return ret;
            }

            private void fillBuffer() throws IOException
            {
                bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
                if (bytesRead == -1)
                    buffer[0] = -1;
            }

            private byte read() throws IOException
            {
                if (bufferPointer == bytesRead)
                    fillBuffer();
                return buffer[bufferPointer++];
            }

            public void close() throws IOException
            {
                if (din == null)
                    return;
                din.close();
            }
        }
        int parent[];
        int rank[];
        public static void main(String[] args) throws IOException {
            sas g = new sas();
            g.go();
        }
        public void go() throws IOException {
            Reader scanner = new Reader();
            int t = scanner.nextInt();
            for (int i = 0; i < t; i++) {
                int n=scanner.nextInt();
                parent=new int[n];
                size=new int[n];
                rank=new int[n];
                parent(n);
                for (int j = 0; j < n; j++) {
                    union(j,scanner.nextInt()-1);
                }
                for (int j = 0; j < n; j++) {
                   printWriter.print(size[find(j)]+" ");
                }
                printWriter.println();
                printWriter.flush();
            }
        }
        public void parent(int n){
            for (int i = 0; i < n; i++) {
                parent[i]=i;
                rank[i]=1;
                size[i]=1;
            }
        }
        public void union(int i,int j){
            int root1=find(i);
            int root2=find(j);
            if(root1==root2){
                return;
            }
            if(rank[root1]>rank[root2]){
                parent[root2]=root1;
                size[root1]+=size[root2];
            }
            else if(rank[root1]<rank[root2]){
                parent[root1]=root2;
                size[root2]+=size[root1];
            }
            else{
                parent[root2]=root1;
                rank[root1]+=1;
                size[root1]+=size[root2];
            }
        }
        public int find(int p){
           if(parent[p]!=p){
               count++;
               parent[p]=find(parent[p]);
           }
           return parent[p];
//            if(parent[p]==-1){
//                return -1;
//            }
//            else if(parent[p]==p){
//                return p;
//            }
//            else {
//                parent[p]=find(parent[p]);
//                return parent[p];
//            }
        }
        public void make(int p){
            parent[p]=p;
            rank[p]=1;
            v[p]=true;
        }
        Random rand = new Random();
        public void sort(int[] a, int n) {
                for (int i = 0; i < n; i++) {
                   int j = rand.nextInt(i + 1);
                   int tmp = a[i];
                   a[i] = a[j];
                   a[j] = tmp;
                }
                Arrays.sort(a, 0, n);
        }
        public long gcd(long a,long b){
               if(b==0){
                  return a;
               }
               return gcd(b,a%b);
        }
        private void dfs2(ArrayList<Integer>[]arrayList,int j,int count){
            ans[j]=count;
            for(int i:arrayList[j]){
                if(ans[i]==-1){
                    dfs2(arrayList,i,count);
                }
            }
        }
        private void dfs(ArrayList<Integer>[] arrayList, int j) {
            v[j]=true;
            count++;
            for(int i:arrayList[j]){
                if(v[i]==false) {
                    dfs(arrayList, i);
                }
            }
        }
}

    class Pair{
    long x;
    long y;
    public Pair(long x,long y){
        this.x=x;
        this.y=y;
    }
}
class Sorting implements Comparator<Pair> {
    public int compare(Pair p1,Pair p2){
        return Long.compare(p1.x,p2.x);
    }

}