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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.PriorityQueue;
import java.util.Scanner;


public class R351C {
	public static void main(String[] args) throws IOException {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		Scanner s = new Scanner(System.in);
		
		int n = Integer.parseInt(f.readLine());
		String[] split = f.readLine().split(" ");
		number[] a = new number[n]; int[] dominant = new int[n], num = new int[n];
		PriorityQueue<number> pq = new PriorityQueue<>();
		for(int i = 0; i < n; i++) {
			num[i] = Integer.parseInt(split[i]);
			a[i] = new number(Integer.parseInt(split[i]));
			if(!pq.contains(a[i]))
				pq.add(a[i]);
		}
		int max=0, maxId=0;
//		System.out.println(Arrays.toString(num));
		HashMap<Integer, Integer> map = new HashMap<>();
		for(int i = 0; i < n; i++) {
			for(int j = i; j < n; j++) {
				if(!map.containsKey(num[j])) {
					map.put(num[j], 1);
				} else {
					map.put(num[j], map.get(num[j])+1);
				}
//				System.out.println(map.toString());
				if(map.get(num[j]) > max) {
//					System.out.println("NEW MAX");
					max = map.get(num[j]);
					maxId=num[j];
					dominant[num[j]-1]++;
				} else if(map.get(num[j]) == max) {
					if(maxId < num[j]) {
						dominant[maxId-1]++;
					} else {
						max = map.get(num[j]);
						maxId=num[j];
						dominant[num[j]-1]++;	
					}
				} else {
					dominant[maxId-1]++;
				}
			}
			max=0;
			map.clear();
//			clear(a);
		}
		for(int i = 0; i < n; i++) System.out.print(dominant[i]+" ");
	}
	
	static void clear(number[] a) {
		for(int i = 0; i < a.length; i++) a[i].oc=0;
	}
}

class number implements Comparable<number> {
	int id, oc, change;
	public number(int id) {
		this.id=id;oc=change=0;
	}
	
	public int compareTo(number one) {
		if(oc==one.oc) return id-one.id;
		return this.oc-one.oc;
	}
	
	public String toString() {return id + " " + oc;}
}