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.*;
import java.util.*;
 
public class bear_colors {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] a = new int[5005];
		for (int i = 1; i <= n; i++)
			a[i] = in.nextInt();
		int[] ans = new int[5005];
		for (int i = 1; i <= 5000; i++)
			ans[i] = 0;
		int[] num = new int[5005];
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= 5000; j++)
				num[j] = 0;
			int maxint = a[i];
			num[maxint]++; 
			ans[maxint]++;
			for (int j = i + 1; j <= n; j++) {
				num[a[j]]++;
				if (num[a[j]] > num[maxint]) 
					maxint = a[j];
				else if (num[a[j]] == num[maxint] && a[j] <maxint)
					maxint = a[j];
				ans[maxint]++;
			}
		}
		for (int i = 1; i <= n; i++)
			System.out.print(ans[i]+" ");

	}
}
