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.util.StringTokenizer;

public class BearAndColors {
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(bf.readLine());
		StringTokenizer st = new StringTokenizer(bf.readLine());
		int a[] = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = Integer.parseInt(st.nextToken());
		int segs[] = new int[n + 1];
		for (int i = 0; i < n; i++) {
			int occs[] = new int[n + 1];
			int max = -1;
			int maxi = -1;
			for (int j = i; j < n; j++) {
				int x = a[j];
				occs[x]++;
				if (occs[x] > max) {
					max = occs[x];
					maxi = x;
					segs[x]++;
				} else if (max == occs[x]) {
					if (maxi < x)
						segs[maxi]++;
					else {
						maxi = x;
						segs[x]++;
					}
				} else {
					if (maxi != -1)
						segs[maxi]++;
				}
			}
		}
		System.out.print(segs[1]);
		for (int i = 2; i <= n; i++)
			System.out.print(" " + segs[i]);
		System.out.println();
	}
}
