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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in Actual solution is at the top
 *
 * @author MakeMoroccoGreatAgain
 */
public class Main {

  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    InputReader in = new InputReader(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    B2BooksExchangeHardVersion solver = new B2BooksExchangeHardVersion();
    solver.solve(1, in, out);
    out.close();
  }

  static class B2BooksExchangeHardVersion {

    public void solve(int testNumber, InputReader in, PrintWriter out) {
      int q = in.nextInt();
      while (q-- > 0) {
        int n = in.nextInt();
        int[] a = in.nextIntArrayOneBased(n);
        int[] ans = new int[n + 1];
        int[] marked = new int[n + 1];

        for (int i = 1; i <= n; i++) {
          if (marked[i] == 0) {
            int size = 0, cur = i;
            do {
              marked[cur] = 1;
              cur = a[cur];
              ++size;
            } while (marked[cur] == 0);
            do {
              marked[cur] = 2;
              ans[cur] = size;
              cur = a[cur];
            } while (marked[cur] != 2);
          }
          out.print(ans[i]);
          out.print(" ");
        }
        out.println();
      }
    }

  }

  static final class InputReader {

    private final InputStream stream;
    private final byte[] buf = new byte[1 << 20];
    private int curChar;
    private int numChars;

    public InputReader() {
      this.stream = System.in;
    }

    public InputReader(final InputStream stream) {
      this.stream = stream;
    }

    private final int read() {
      if (this.numChars == -1) {
        throw new UnknownError();
      } else {
        if (this.curChar >= this.numChars) {
          this.curChar = 0;

          try {
            this.numChars = this.stream.read(this.buf);
          } catch (IOException ex) {
            throw new InputMismatchException();
          }

          if (this.numChars <= 0) {
            return -1;
          }
        }

        return this.buf[this.curChar++];
      }
    }

    public final int nextInt() {
      int c;
      for (c = this.read(); isSpaceChar(c); c = this.read()) {
      }

      byte sgn = 1;
      if (c == 45) { // 45 == '-'
        sgn = -1;
        c = this.read();
      }

      int res = 0;

      while (c >= 48 && c <= 57) { // 48 == '0', 57 == '9'
        res *= 10;
        res += c - 48; // 48 == '0'
        c = this.read();
        if (isSpaceChar(c)) {
          return res * sgn;
        }
      }

      throw new InputMismatchException();
    }

    private static final boolean isSpaceChar(final int c) {
      return c == 32 || c == 10 || c == 13 || c == 9
          || c == -1; // 32 == ' ', 10 == '\n', 13 == '\r', 9 == '\t'
    }

    public final int[] nextIntArrayOneBased(final int n) {
      int[] ret = new int[n + 1];
      for (int i = 1; i <= n; i++) {
        ret[i] = nextInt();
      }
      return ret;
    }

  }
}