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.Scanner;

/* Pravim 3 niza. Na pocetku ucitam n, i niz1. 
 * Zatim niz2 koristim za sve podnizove niza1, 
 * i u niz3 upisujem, na pozicije brojeva koji su dominantni u nizu,
 *  vrednost koliko puta su se oni javili kao dominantni u tim podnizovima.
 *  Slozenost algoritma je O(n + n + n*n), odnosno O(n^2)*/

public class MainClass {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k;
		int[] niz1=new int[n];
		int[] niz2 = new int[n+1];
		int[] niz3 = new int[n+1];
		for(int i=0;i<n;i++){
			niz1[i] = sc.nextInt();
		}
		sc.close();
		for(int i=1;i<n+1;i++){
			niz3[i]=0;
		}
		int max_tren=0;
		for(int i=0;i<n;i++){
			for(int j=0;j<n+1;j++) niz2[j]=0;
			max_tren=niz1[i];
	        niz2[niz1[i]]++;
	        niz3[niz1[i]]++;
	        k=i+1;
	        while(k < n){
	            niz2[niz1[k]]++;
	            if(niz2[niz1[k]] > niz2[max_tren]){
	                max_tren=niz1[k];
	                niz3[niz1[k]]++;
	            }else {
	                if(niz2[niz1[k]]==niz2[max_tren]){
	                    if(niz1[k]<max_tren){
	                        max_tren=niz1[k];
	                        niz3[niz1[k]]++;
	                    }else{
	                        niz3[max_tren]++;
	                    }
	                }else{
	                    niz3[max_tren]++;
	                }
	            }
	            k++;

	        }
		}
		for(int i=1;i<n+1;i++){
			System.out.print(niz3[i]+" ");
		}

	}

}
  		 	 	 				 			     		 				