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.lang.*;
import java.io.*;

public class Main {

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String line;
        String[] ar;
        StringBuilder sb = new StringBuilder();
        //-----------------------------------------------------------------------
        int n = Integer.parseInt(br.readLine());
        ar = br.readLine().split(" ");
        int[] a = new int[n];
        int[] sol = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(ar[i]) - 1;
        }
        for (int i = 0; i < n; i++) {
            int[] f = new int[n];
            f[a[i]]++;
            int best = 1, besti = a[i];
            sol[besti]++;
            for (int j = i + 1; j < n; j++) {
                f[a[j]]++;
                if (f[a[j]] > best) {
                    best = f[a[j]];
                    besti = a[j];
                } else if (f[a[j]] == best && a[j] < besti) {
                    besti = a[j];
                }
                sol[besti]++;
            }
        }
        for (int i = 0; i < n; i++) {
            sb.append(sol[i] + " ");
        }
        //-----------------------------------------------------------------------
        System.out.println(sb);
    }
}
