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.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.*; 
import java.util.StringTokenizer; 
import java.io.PrintWriter;
import java.io.*;
import java.util.stream.Collectors.*;
import java.lang.*;
import static java.util.stream.Collectors.*;
import static java.util.Map.Entry.*;
public class Ideo
{ 
    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }
   
static int power(int x,int y) 
{ 
    int res = 1;      // Initialize result 
  
      // Update x if it is more than or  
                // equal to p 
  
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if (y%2==1) 
            res = (res*x); 
  
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x);   
    } 
    return res; 
} 
 
 static long gcd(long a,long b) 
    { 
        if (a == 0) 
            return b; 
        return gcd(b % a, a); 
    } 
 
 
  static boolean compareSeq(char[] S, int x, int y, int n) 
    { 
        for (int i = 0; i < n; i++)
         {
 
            if (S[x] < S[y]) 
               return true;
            else if (S[x] > S[y]) 
                return false; 
            x = (x + 1) % n; 
            y = (y + 1) % n; 
        } 
        return true; 
    } 
 
    static void build(long[] sum,int[] arr,int n)
    {
      for(int i=0;i<(1<<n);i++)
      {
        long total=0;
        for(int j=0;j<n;j++)
        {
          if((i & (1 << j)) > 0)
            total+=arr[j];
        }
 
        sum[i]=total;
      }
    }
 
 
     static int count(long arr[], long x, int n) 
       {
 
        int l=0;
        int h=n-1;
 
        int res=-1;
        int mid=-1;
        while(l<=h)
        {
          mid=l+(h-l)/2;
          if(x==arr[mid])
          {
            res=mid;
            h=mid-1;
          }
          else if(x<arr[mid])
            h=mid-1;
          else
            l=mid+1;
        }
 
        if(res==-1)
          return 0;
 
        //res is first index and res1 is last index of an element in a sorted array total number of occurences is (res1-res+1)
 
        int res1=-1;
        l=0;
        h=n-1;
        while(l<=h)
        {
          mid=l+(h-l)/2;
          if(x==arr[mid])
          {
            res1=mid;
            l=mid+1;
          }
          else if(x<arr[mid])
            h=mid-1;
          else
            l=mid+1;
        }
 
 
 
        if(res1==-1)
          return 0;
        if(res!=-1 && res1!=-1)
          return (res1-res+1);
 
        return 0;
 
       } 
 
       static int parity(int a)
       {
    a^=a>>16;
    a^=a>>8;
    a^=a>>4;
    a^=a>>2;
    a^=a>>1;
    return a&1;
  }
/*
  PriorityQueue<aksh> pq = new PriorityQueue<>((o1, o2) -> {
            if (o1.p < o2.p)
                return 1;
            else if (o1.p > o2.p)
                return -1;
            else
                return 0;
 
        });//decreasing order acc to p*/
static int power(int x, int y, int m)  
    { 
        if (y == 0) 
            return 1; 
          
        int p = power(x, y / 2, m) % m; 
        p = (p * p) % m; 
      
        if (y % 2 == 0) 
            return p; 
        else
            return (x * p) % m; 
    } 
/*static int modinv(int a, int m) 
    { 
        int g = gcd(a, m); 
        if (g != 1) 
            return 0;
        else 
        { 
           return power(a, m - 2, m); 
        } 
        //return 0;
    } */
 
static int[] product(int[] nums) {
    int[] result = new int[nums.length];
 
    int[] t1 = new int[nums.length];
    int[] t2 = new int[nums.length];
 
    t1[0]=1;
    t2[nums.length-1]=1;
 
    //scan from left to right
    for(int i=0; i<nums.length-1; i++){
        t1[i+1] = nums[i] * t1[i];
    }
 
    //scan from right to left
    for(int i=nums.length-1; i>0; i--){
        t2[i-1] = t2[i] * nums[i];
    }
 
    for(int i=0;i<nums.length;i++)
      {
        System.out.print(t1[i]+" "+t2[i]);
        System.out.println();
      }
 
    //multiply
    for(int i=0; i<nums.length; i++){
        result[i] = t1[i] * t2[i];
    }
 
    return result;
}
 
static int getsum(int[] bit,int ind)
{
  int sum=0;
  while(ind>0)
  {
    sum+=bit[ind];
    ind-= ind & (-ind);
  }
 
  return sum;
}
 
static void update(int[] bit,int max,int ind,int val)
{
  while(ind<=max)
  {
    bit[ind]+=val;
    ind+= ind & (-ind);
  }
}
 
    //static ArrayList<Integer>[] adj;
 
    static boolean check(long mid,long a,long b)
    {
      long count=1;
      while(count<=mid)
      {
        count++;
        if(a<b)
          a+=count;
        else
          b+=count;
 
        if(a==b)
          return true;
      }
        return false;
    }
 
    static class aksh implements Comparable<aksh>
    {
      int p;
      int q;
 
      public aksh(int p,int q)
      {
        this.p=p;
        this.q=q;
      }
 
      public int compareTo(aksh o)
      {
        if(p!=o.p)
          return p-o.p;
        else
          return q-o.p;
      }
    }
    static int get(int arr[], int n) 
    { 
        int result = 0; 
        int x, sum; 
          
        // Iterate through every bit 
        for(int i=0; i<32; i++) 
        { 
            // Find sum of set bits at ith position in all  
            // array elements 
            sum = 0; 
            x = (1 << i); 
            for(int j=0; j<n; j++) 
            { 
                if((arr[j] & x)!=0) 
                sum++; 
            } 
            // The bits with sum not multiple of 3, are the 
            // bits of element with single occurrence. 
            if ((sum % 3)!=0) 
            result |= x; 
        } 
        return result; 
    } 
    /*        Collections.sort(orders, new Comparator<Point>() {
            @Override
            public int compare(Point o1, Point o2) {
                if (o1.diff < o2.diff) {
                    return -1;
                } else if (o1.diff > o2.diff) {
                    return 1;
                } else {
                    return 0;
                }
            }
        });  */
 
 
    static boolean isPrime(int n)  
    {  
        // Corner cases  
        if (n <= 1) return false;  
        if (n <= 3) return true;  
          
        // This is checked so that we can skip  
        // middle five numbers in below loop  
        if (n % 2 == 0 || n % 3 == 0) return false;  
          
        for (int i = 5; i * i <= n; i = i + 6)  
            if (n % i == 0 || n % (i + 2) == 0)  
            return false;  
          
        return true;  
    }  
      
    static int nextPrime(int N)  
    {  
      
        // Base case  
        if (N <= 1)  
            return (2-N);  
      
        int prime = N;  
        boolean found = false;  
      
        // Loop continuously until isPrime returns  
        // true for a number greater than n  
        while (!found)  
        {  
            if (isPrime(prime))  
                {
                  found = true;
                  break;
                }  
 
            prime++;
        }  
      
        return (prime-N);  
    }  
 
 
    static long product(long x) 
{ 
    long prod = 1; 
    while (x > 0)  
    { 
        prod *= (x % 10); 
        x /= 10; 
    } 
    return prod; 
} 
  
// This function returns the number having 
// maximum product of the digits 
static long findNumber(long l, long r) 
{ 
    // Converting both integers to strings 
    //string a = l.ToString(); 
    String b = Long.toString(r); 
  
    // Let the current answer be r 
    long ans = r; 
    for (int i = 0; i < b.length(); i++)  
    { 
        if (b.charAt(i) == '0') 
            continue; 
  
        // Stores the current number having 
        // current digit one less than current 
        // digit in b 
        char[] curr = b.toCharArray(); 
        curr[i] = (char)(((int)(curr[i] -  
                    (int)'0') - 1) + (int)('0')); 
  
        // Replace all following digits with 9 
        // to maximise the product 
        for (int j = i + 1; j < curr.length; j++) 
            curr[j] = '9'; 
  
        // Convert string to number 
        int num = 0; 
        for (int j = 0; j < curr.length; j++) 
            num = num * 10 + (curr[j] - '0'); 
  
        // Check if it lies in range and its product 
        // is greater than max product 
        if (num >= l && product(ans) < product(num)) 
            ans = num; 
    } 
  
    return product(ans); 
} 
 
static long mod=998244353;
 
static long pow(long in, long pow) {
    if(pow == 0) return 1;
    long out = pow(in, pow / 2);
    out = (out * out) % mod;
    if(pow % 2 == 1) out = (out * in) % mod;
    return out;
  }
  static long inv(long in) {
    return pow(in, mod - 2);
  }
 
  static void swap(int x,int y)
  {
    int temp=x;
    x=y;
    y=temp;
  }
 
  static int[] par;
  static int[] size;
  static int find(int i) 
  {
     if (par[i] == i)
      return i;
 
    return par[i] = find(par[par[i]]);
  }
 
  static void union(int x, int y) 
  {
      x = find(x);
      y = find(y);
      if (x == y)
        return;
      if (size[x] < size[y])
        swap(x, y);
 
      par[y] = x;
      size[x] += size[y];
    }
 
 
    static void multisourcebfs(long[] arr,int n,int m)
    {
      //HashSet<Long> vis=new HashSet<>();
      HashMap<Long,Long> dis=new HashMap<>();
      Queue<Long> q=new LinkedList<>();
 
      for(int i=0;i<arr.length;i++)
      {
        dis.put(arr[i],(long)0);
        q.add(arr[i]);
        //vis.add(arr[i]);
      }
 
      long[] res=new long[m];
      long ans=0;
      int k=0;
      while(!q.isEmpty())
      {
        if(k==m)
          break;
        long x=q.remove();
        if(dis.get(x)!=0)
        {
          ans+=dis.get(x);
          res[k]=x;
          k++;
        }
 
        if(dis.get(x-1)==null)
        {
          dis.put(x-1, dis.get(x)+1);
          q.add(x-1);
          //vis.add(x-1);
        }
 
        if(dis.get(x+1)==null)
        {
          dis.put(x+1, dis.get(x)+1);
          q.add(x+1);
          //vis.add(x+1);
        }
      }
 
      System.out.println(ans);
 
      for(int i=0;i<m;i++)
        System.out.print(res[i]+" ");
 
    }
 
    static int x; 
    static int maxcount=0;
    static void dfs(int node,int par,int dist)
    {
        if(dist>=maxcount)
        {
           maxcount=dist;
           x=node;
        }
 
        List<Integer> l = adj[node]; 
        for(Integer i: l)
        {
          if(i!=par)
          dfs(i,node,dist+1);
        }
    }

    static boolean func(int mid, aksh[] a, int n)
    {
      for(int i=0;i<n;i++)
      {
        //System.out.println(mid);
        int x=a[i].p;
        int y=a[i].p;

        if(a[i].q!=0)
          x=(int)(a[i].p/a[i].q);
        if((n-a[i].q-1)!=0)
           y=(int)(a[i].p/(n-a[i].q-1));
        int temp=Math.min(x,y);
         if(temp<mid)
          return false;
      }
      return true;
    }

    static int h;
    static int w;
    static int bfs(char[][] ch,int i,int j)
    {
        Queue<aksh> q=new LinkedList<>();
        q.add(new aksh(i,j));

        int[][] dis=new int[h][w];
        int[][] vis=new int[h][w];
        vis[i][j]=1;
        dis[i][j]=0;
 
        while(!q.isEmpty())
        {
          aksh arr=q.poll();
          int x1=arr.p;
          int y1=arr.q;
 
          if((x1-1)>=0 && ch[x1-1][y1]!='#' && vis[x1-1][y1]==0)
          {
            vis[x1-1][y1]=1;
            q.add(new aksh(x1-1, y1));
            dis[x1-1][y1]=Math.max(dis[x1-1][y1],dis[x1][y1]+1);
          }
 
          if((x1+1)<h && ch[x1+1][y1]!='#' && vis[x1+1][y1]==0)
          {
            vis[x1+1][y1]=1;
            dis[x1+1][y1]=Math.max(dis[x1+1][y1],dis[x1][y1]+1);
            q.add(new aksh(x1+1, y1));
          }
 
          if((y1-1)>=0 && ch[x1][y1-1]!='#' && vis[x1][y1-1]==0)
          {
            vis[x1][y1-1]=1;
            dis[x1][y1-1]=Math.max(dis[x1][y1-1],dis[x1][y1]+1);
            q.add(new aksh(x1,y1-1));
          }
 
    
          if((y1+1)<w && ch[x1][y1+1]!='#' && vis[x1][y1+1]==0)
          {
            vis[x1][y1+1]=1;
            dis[x1][y1+1]=Math.max(dis[x1][y1+1],dis[x1][y1]+1);
            q.add(new aksh(x1,y1+1));
          }
        }
        int max=Integer.MIN_VALUE;

        for(int x=0;x<h;x++)
        {
          for(int y=0;y<w;y++)
            max=Math.max(dis[x][y],max);
        }

        return max;
    }

    static final long INF = Long.MAX_VALUE/5;
    static ArrayList<Integer>[] adj;
    public static void main(String args[] ) throws Exception 
    {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT */
 
        FastReader sc=new FastReader();
        int t=sc.nextInt();

        while(t-->0)
        {
          int n=sc.nextInt();
          int[] p=new int[n+1];

          for(int i=1;i<=n;i++)
            p[i]=sc.nextInt();

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

          for(int i=1;i<=n;i++)
          {
            if(vis[i]==0)
            {
              int j=i;
              int c=1;
              ArrayList<Integer> arr=new ArrayList<>();
              while(p[j]!=i)
              {
                arr.add(p[j]);
                j=p[j];
                vis[j]=1;
                c++;
              }

              vis[i]=1;
              ans[i]=c;

              for(int x=0;x<arr.size();x++)
                ans[arr.get(x)]=c;
            }
          }

          for(int i=1;i<=n;i++)
            System.out.print(ans[i]+" ");





        }
       
        
        }
 
    }