Bear Limak has n colored balls, arranged in one long row. Balls are numbered 1 through n, from left to right. There are n possible colors, also numbered 1 through n. The i-th ball has color ti.

For a fixed interval (set of consecutive elements) of balls we can define a dominant color. It's a color occurring the biggest number of times in the interval. In case of a tie between some colors, the one with the smallest number (index) is chosen as dominant.

There are <image> non-empty intervals in total. For each color, your task is to count the number of intervals in which this color is dominant.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 5000) — the number of balls.

The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ n) where ti is the color of the i-th ball.

Output

Print n integers. The i-th of them should be equal to the number of intervals where i is a dominant color.

Examples

Input

4
1 2 1 2


Output

7 3 0 0 


Input

3
1 1 1


Output

6 0 0 

Note

In the first sample, color 2 is dominant in three intervals:

  * An interval [2, 2] contains one ball. This ball's color is 2 so it's clearly a dominant color. 
  * An interval [4, 4] contains one ball, with color 2 again. 
  * An interval [2, 4] contains two balls of color 2 and one ball of color 1. 



There are 7 more intervals and color 1 is dominant in all of them.
import java.util.*;
import java.io.*;
public class BearandColors
{


    /************************ SOLUTION STARTS HERE ***********************/


    private static void solve(FastScanner s1, PrintWriter out){

	int N = s1.nextInt();
	int arr[]  = s1.nextIntArray(N);
	int cnt[]  = new int[N + 1];
	int freq[] = new int[N + 1];
	for(int i=0;i<N;i++){	    
	    Arrays.fill(freq, 0);
	    int best = arr[i];
	    freq[best]++;
	    cnt[best]++;
	    for(int j=i+1;j<N;j++){
		freq[arr[j]]++;
		best = ((freq[arr[j]] > freq[best]) || (freq[arr[j]] == freq[best] && arr[j] < best)) ? arr[j] : best;
		cnt[best]++;
	    }
	}
	for(int i=1;i<=N;i++)
	    out.print(cnt[i] + " ");
    }



    /************************ SOLUTION ENDS HERE ************************/






    /************************ TEMPLATE STARTS HERE *********************/

    public static void main(String []args) throws IOException {
	FastScanner in  = new FastScanner(System.in);
	PrintWriter out = 
		new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false); 
	solve(in, out);
	in.close();
	out.close();
    }    

    static class FastScanner{
	BufferedReader reader;
	StringTokenizer st;
	FastScanner(InputStream stream){reader=new BufferedReader(new InputStreamReader(stream));st=null;}	
	String next()
	{while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}		    
	st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
	String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}	    	  	
	int    nextInt()   {return Integer.parseInt(next());}
	long   nextLong()  {return Long.parseLong(next());}		
	double nextDouble(){return Double.parseDouble(next());}
	char   nextChar()  {return next().charAt(0);}
	int[]  nextIntArray(int n)         {int[] arr= new int[n];   int i=0;while(i<n){arr[i++]=nextInt();}  return arr;}
	long[] nextLongArray(int n)        {long[]arr= new long[n];  int i=0;while(i<n){arr[i++]=nextLong();} return arr;}	
	int[]  nextIntArrayOneBased(int n) {int[] arr= new int[n+1]; int i=1;while(i<=n){arr[i++]=nextInt();} return arr;}	    	
	long[] nextLongArrayOneBased(int n){long[]arr= new long[n+1];int i=1;while(i<=n){arr[i++]=nextLong();}return arr;}	    	
	void   close(){try{reader.close();}catch(IOException e){e.printStackTrace();}}				
    }

    /************************ TEMPLATE ENDS HERE ************************/
}