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.*;
public final class BearColors
{
	public static void main(String[] args) {
		Scanner br=new Scanner(System.in);
		int n=br.nextInt();
		int[] a=new int[5010];
		for(int i=0;i<n;i++)
			a[i]=br.nextInt();
		int[] count=new int[5010];
		int[] c;
		int max=Integer.MIN_VALUE,prev=Integer.MAX_VALUE;
		for(int i=0;i<n;i++)
		{
			c=new int[5010];
			max=Integer.MIN_VALUE;
			prev=Integer.MAX_VALUE;
			for(int j=i;j<n;j++)
			{
				c[a[j]]++;
				if(c[a[j]]>max)
				{
					max=c[a[j]];
					count[a[j]]++;
					prev=a[j];
				}
				else if(c[a[j]]==max)
				{
					if(a[j]<prev)
					{
					count[a[j]]++;
					prev=a[j];
					}
					else
					{
						count[prev]++;
					}

				}
				else if(c[a[j]]<max)
					count[prev]++;
						

			}
		}
		for(int k=1;k<=n;k++)
			System.out.print(count[k]+" ");
		

	}
}