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.math.BigInteger;
import java.util.*;



public class Main {

    private static Set<List<Integer>> permutations;

    public static void main(String[] args) throws IOException {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        InputReader.OutputWriter out = new InputReader.OutputWriter(outputStream);
        Scanner scanner = new Scanner(System.in);

        int t = in.nextInt();
        while (t-->0) {
            int n = in.nextInt();
            int [] a = new int[n];
            for (int i = 0; i < a.length; i++) {
                a[i] = in.nextInt();
                a[i]--;
            }

            boolean[] used = new boolean[n];
            int [] ans = new int[n];
            for (int i = 0; i < a.length; i++) {
                if(!used[i]) {
                    List<Integer> list = new ArrayList<>();
                    while (!used[i]) {
                        list.add(i);
                        used[i] = true;
                        i = a[i];
                    }
                    for (int l : list) {
                        ans[l] = list.size();
                    }
                }
            }
            for (int i = 0; i < n; i++) {
                out.print(ans[i] + " ");
            }
            out.println();
        }

        /*int t = in.nextInt();
        while (t-->0) {
            long n = in.nextLong();
            String s = stepen(n);
            StringBuilder sb = new StringBuilder();
            boolean found = false;
            for (int i = s.length() - 1;i>=0;i--) {
                char cur = s.charAt(i);
                if(cur == '2') {
                    sb.append('0');
                    found = true;
                }
                else {
                    if(found) {
                        if(cur == '0') {
                            found = false;
                        }
                        sb.append('0');
                    }
                    else {
                        sb.append(s.charAt(i));
                        found = false;
                    }
                }
            }
            if(found) {
                sb.append('1');
            }
            out.println(s);
            String ss = sb.reverse().toString();
            out.println(ss);
            out.println(convert(ss));

        }*/

        out.flush();
    }
    private static long convert(String s) {
        char [] c = s.toCharArray();
        long ans = 0;
        int n = c.length;
        for (int i = c.length - 1; i >=0; i--) {
            long value = c[i] - '0';
            ans+=(value * (long) Math.pow(3,(n-i-1)));
        }
        return ans;
    }
    private static String stepen(long n) {
        StringBuilder s = new StringBuilder();
        while (n > 0) {
            s.append(n%3);
            n/=3;
        }
        return s.reverse().toString();
    }
    private static boolean isGood(long n) {
        while (n > 0) {
            long k = n%3;
            n/=3;
            if(k > 1) return false;
        }
        return true;
    }
    private static boolean [] erath() {
        boolean[] isPrime = new boolean[200_000 + 1];
        Arrays.fill(isPrime,true);
        isPrime[0] = false;
        isPrime[1] = false;
        int n = isPrime.length - 1;
        for (int i = 2; i*i<=n; i++) {
            if(isPrime[i]) {
                for (int j = i*i; j <=n; j+=i) {
                    isPrime[j] = false;
                }
            }
        }
        return isPrime;
    }



    public static long gcd(long a, long b) {
       while (b > 0) {
           long temp = a;
           a = b;
           b = temp%b;
       }
       return a;
    }


    private static int factorization(long n) {
        Set<Long> set = new HashSet<>();
        for(long i = 2; i * i <=n; i++) {
            while (n % i == 0) {
                set.add(i);
                n /= i;
            }
        }
        if(n > 1) set.add(n);
        return set.size();
    }

    private static void permute(int [] a, int l, int r) {
        if(l == r) {
            List<Integer> list = new ArrayList<>();
            for(int i : a) list.add(i);
            permutations.add(list);
        }
        else {
            for (int i = l; i <=r; i++) {
                swap(a,l,i);
                permute(a,l+1,r);
                swap(a,l,i);
            }
        }
    }
    private static void swap(int [] a, int l, int r) {
        int temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }

}

class InputReader extends BufferedReader {
    StringTokenizer tokenizer;

    public InputReader(InputStream inputStream) {
        super(new InputStreamReader(inputStream), 32768);
    }

    public InputReader(String filename) {
        super(new InputStreamReader(Thread.currentThread().getContextClassLoader().getResourceAsStream(filename)));
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(readLine());
            } catch (IOException e) {
                throw new RuntimeException();
            }
        }
        return tokenizer.nextToken();
    }

    public Integer nextInt(){
        return Integer.valueOf(next());
    }
    public Long nextLong() {
        return  Long.valueOf(next());
    }
    public Double nextDouble() {
        return Double.valueOf(next());
    }
    static class OutputWriter extends PrintWriter {
        public OutputWriter(OutputStream outputStream) {
            super(outputStream);
        }

        public OutputWriter(Writer writer) {
            super(writer);
        }

        public OutputWriter(String filename) throws FileNotFoundException {
            super(filename);
        }

        public void close() {
            super.close();
        }
    }
}