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.*;

public class Rough3 implements Runnable {

	MyFastIOptReader scn;
	PrintWriter syso;
	boolean judge = true;
	String INPUT = "";
	static int mod = 987654319;

	public static void main(String[] args) {
		new Thread(null, new Rough3(), "Main", 1 << 26).start();
	}

	void Solution() {

		int T = scn.nextInt();
		StringBuilder sb = new StringBuilder();
		while (T-- > 0) {
			int n = scn.nextInt();
			int ar[] = new int[n + 1];
			for (int i = 1; i <= n; i++) {
				ar[i] = scn.nextInt();

			}
			int ans[] = new int[n + 1];

			for (int i = 1; i <= n; i++) {
				if (ans[i] != 0)
					continue;
				int go = ar[i];
				int gg = 1;
				while (go != i) {
					go = ar[go];
					gg++;
				}
				int x = i;
				for (int k = 0; k < gg; k++) {
					ans[x] = gg;
					x = ar[x];
				}
			}
			for (int i = 1; i <= n; i++)
				sb.append(ans[i] + " ");
			sb.append("\n");
		}
		syso.print(sb);
	}

	public void run() {
		long time = System.currentTimeMillis();
		boolean oj = System.getProperty("ONLINE_JUDGE") != null || judge;
		syso = new PrintWriter(System.out);
		scn = new MyFastIOptReader(oj);
		Solution();
		syso.flush();
		if (!oj) {
			System.out.println(Arrays.deepToString(new Object[] { System.currentTimeMillis() - time + " ms" }));
		}
	}

	// Taking Fast IO from Internet

	class MyFastIOptReader {

		InputStream is;

		public MyFastIOptReader(boolean onlineJudge) {
			is = onlineJudge ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		}

		byte[] inbuf = new byte[1024];
		public int lenbuf = 0, ptrbuf = 0;

		int readByte() {
			if (lenbuf == -1)
				throw new InputMismatchException();
			if (ptrbuf >= lenbuf) {
				ptrbuf = 0;
				try {
					lenbuf = is.read(inbuf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (lenbuf <= 0)
					return -1;
			}
			return inbuf[ptrbuf++];
		}

		boolean isSpaceChar(int c) {
			return !(c >= 33 && c <= 126);
		}

		int skip() {
			int b;
			while ((b = readByte()) != -1 && isSpaceChar(b))
				;
			return b;
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		char nextChar() {
			return (char) skip();
		}

		String next() {
			int b = skip();
			StringBuilder sb = new StringBuilder();
			while (!(isSpaceChar(b))) {
				sb.appendCodePoint(b);
				b = readByte();
			}
			return sb.toString();
		}

		String nextLine() {
			int b = skip();
			StringBuilder sb = new StringBuilder();
			while ((!isSpaceChar(b) || b == ' ')) {
				sb.appendCodePoint(b);
				b = readByte();
			}
			return sb.toString();
		}

		char[] next(int n) {
			char[] buf = new char[n];
			int b = skip(), p = 0;
			while (p < n && !(isSpaceChar(b))) {
				buf[p++] = (char) b;
				b = readByte();
			}
			return n == p ? buf : Arrays.copyOf(buf, p);
		}

		int nextInt() {
			int num = 0, b;
			boolean minus = false;
			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
				;
			if (b == '-') {
				minus = true;
				b = readByte();
			}

			while (true) {
				if (b >= '0' && b <= '9') {
					num = num * 10 + (b - '0');
				} else {
					return minus ? -num : num;
				}
				b = readByte();
			}
		}

		int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		long nextLong() {

			long num = 0;
			int b;
			boolean minus = false;
			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
				;
			if (b == '-') {
				minus = true;
				b = readByte();
			}
			while (true) {
				if (b >= '0' && b <= '9') {
					num = num * 10 + (b - '0');
				} else {
					return minus ? -num : num;
				}
				b = readByte();
			}
		}
	}
}