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.awt.Point;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;


public class Ds{
	
	public static void main(String[] args) throws IOException {
		
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		
		int n = Integer.parseInt(in.readLine());
		
		String[] line = in.readLine().split("\\s+");
		
		int[] arr = new int[n];
		
		
		for(int i=0; i<n; i++) {
			arr[i] = Integer.parseInt(line[i]) - 1;
		}
		
		int[] colors = new int[n];
		
		for(int i=0; i<n; i++) {
			
			int[] freq = new int[n+1];
			
			Point p = new Point(n, 0);
			
			for(int j=i; j<n; j++) {
				
				freq[arr[j]]++;
				
				if(freq[arr[j]] > p.y || (freq[arr[j]] == p.y && arr[j] < p.x)) {
					p.y = freq[arr[j]];
					p.x = arr[j];
				}
				
				colors[p.x]++;
			}
		}
		
		for(int i : colors) {
			System.out.print(i + " ");
		}
		
	
		
		
	}
	
}