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.io.BufferedReader;
import java.io.IOException;
import java.util.Collections;
import java.io.InputStreamReader;


public class Main1 {




    public static void main(String[] args) {

        try {
            FastScanner in = new FastScanner();
            PrintWriter out = new PrintWriter(System.out);


            int t = in.nextInt() ;
            while (t-- > 0){
                int n= in.nextInt() ;
                int a[] = new int[n] ;

                for (int i = 0; i <n ; i++) {
                    a[i] = in.nextInt()-1 ;
                }


                int ans[] = new int[n] ;


            boolean vis[] = new boolean[n] ;

            for (int i = 0; i < n; i++) {
                if (vis[i]) continue;

                int curr = i ;
                int cnt = 0 ;

                while (!vis[curr]){
                    vis[curr] = true ;
                    cnt++ ;
                    curr = a[curr] ;
                }
                int temp = cnt ;
                int tt = curr ;
                while (temp-- > 0){
                    ans[tt] = cnt ;
                    tt = a[tt] ;

                }

            }

            for (int i : ans)
                out.append( i  + " ") ;
            out.append("\n") ;
            }







            out.flush();
            out.close();
        } catch (Exception e) {
            return;
        }

    }


    static int findLCA(int a, int b, int par[][], int depth[]) {
        if (depth[a] > depth[b]) {
            a = a ^ b;
            b = a ^ b;
            a = a ^ b;
        }
        int diff = depth[b] - depth[a];
        for (int i = 19; i >= 0; i--) {
            if ((diff & (1 << i)) > 0) {
                b = par[b][i];
            }
        }
        if (a == b)
            return a;
        for (int i = 19; i >= 0; i--) {
            if (par[b][i] != par[a][i]) {
                b = par[b][i];
                a = par[a][i];
            }
        }
        return par[a][0];
    }

    static int gcd(int a, int b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static void formArrayForBinaryLifting(int n, int par[][]) {
        for (int j = 1; j < 20; j++) {
            for (int i = 0; i < n; i++) {
                if (par[i][j - 1] == -1)
                    continue;
                par[i][j] = par[par[i][j - 1]][j - 1];
            }
        }
    }


    static void sort(int ar[]) {
        int n = ar.length;
        ArrayList<Integer> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static void sort1(long ar[]) {
        int n = ar.length;
        ArrayList<Long> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static long ncr(long n, long r, long mod) {
        if (r == 0)
            return 1;
        long val = ncr(n - 1, r - 1, mod);
        val = (n * val) % mod;
        val = (val * modInverse(r, mod)) % mod;
        return val;
    }

    static long fast_pow(long base, long n, long M) {
        if (n == 0)
            return 1;
        if (n == 1)
            return base % M;
        long halfn = fast_pow(base, n / 2, M);
        if (n % 2 == 0)
            return (halfn * halfn) % M;
        else
            return (((halfn * halfn) % M) * base) % M;
    }

    static long modInverse(long n, long M) {
        return fast_pow(n, M - 2, M);
    }


    private static int countDigits(long l) {
        if (l >= 1000000000000000000L) return 19;
        if (l >= 100000000000000000L) return 18;
        if (l >= 10000000000000000L) return 17;
        if (l >= 1000000000000000L) return 16;
        if (l >= 100000000000000L) return 15;
        if (l >= 10000000000000L) return 14;
        if (l >= 1000000000000L) return 13;
        if (l >= 100000000000L) return 12;
        if (l >= 10000000000L) return 11;
        if (l >= 1000000000L) return 10;
        if (l >= 100000000L) return 9;
        if (l >= 10000000L) return 8;
        if (l >= 1000000L) return 7;
        if (l >= 100000L) return 6;
        if (l >= 10000L) return 5;
        if (l >= 1000L) return 4;
        if (l >= 100L) return 3;
        if (l >= 10L) return 2;
        return 1;
    }


    static class FastOutput implements AutoCloseable, Closeable, Appendable {
        private static final int THRESHOLD = 32 << 10;
        private final Writer os;
        private StringBuilder cache = new StringBuilder(THRESHOLD * 2);

        public FastOutput append(CharSequence csq) {
            cache.append(csq);
            return this;
        }

        public FastOutput append(CharSequence csq, int start, int end) {
            cache.append(csq, start, end);
            return this;
        }

        private void afterWrite() {
            if (cache.length() < THRESHOLD) {
                return;
            }
            flush();
        }

        public FastOutput(Writer os) {
            this.os = os;
        }

        public FastOutput(OutputStream os) {
            this(new OutputStreamWriter(os));
        }

        public FastOutput append(char c) {
            cache.append(c);
            afterWrite();
            return this;
        }

        public FastOutput append(long c) {
            cache.append(c);
            afterWrite();
            return this;
        }

        public FastOutput append(String c) {
            cache.append(c);
            afterWrite();
            return this;
        }

        public FastOutput println() {
            return append(System.lineSeparator());
        }

        public FastOutput flush() {
            try {
                //            boolean success = false;
                //            if (stringBuilderValueField != null) {
                //                try {
                //                    char[] value = (char[]) stringBuilderValueField.get(cache);
                //                    os.write(value, 0, cache.length());
                //                    success = true;
                //                } catch (Exception e) {
                //                }
                //            }
                //            if (!success) {
                os.append(cache);
                //            }
                os.flush();
                cache.setLength(0);
            } catch (IOException e) {
                throw new UncheckedIOException(e);
            }
            return this;
        }

        public void close() {
            flush();
            try {
                os.close();
            } catch (IOException e) {
                throw new UncheckedIOException(e);
            }
        }

        public String toString() {
            return cache.toString();
        }

    }


    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 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 nextLong() {
            return Long.parseLong(next());
        }
    }


}