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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class B {

	public static void main(String[] args) throws IOException{
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n = sc.nextInt();
		int[] a = new int[n]; for(int i = 0; i < n; i++)a[i] = sc.nextInt();
		Point[][] ans = new Point[n][];
		ans[0] = new Point[1];
		ans[0][0] = new Point(a[0], 1);
		for(int i = 1; i < n; i++)
		{
			int size = i + 1;
			ans[i] = new Point[size];
			ans[i][0] = new Point(a[i], 1);
			int curOcc = 1;
			for(int j = i-1; j >= 0; j--)
			{
				if(a[j] == a[i]) curOcc++;
				int lastOcc = ans[i - 1][i - j - 1].y, last = ans[i - 1][i - j - 1].x;
				if(curOcc > lastOcc || curOcc == lastOcc && a[i] <= last)
					ans[i][i - j] = new Point(a[i], curOcc);
				else
					ans[i][i - j] = new Point(last, lastOcc);
			}
		}
		long[] occ = new long[n+1];
		for(int i = 0; i < n; i++)
			for(int j = 0; j < ans[i].length; j++)
				occ[ans[i][j].x]++;
		for(int i = 1; i <= n; i++)
			out.print(occ[i] + " ");
		out.flush();
	}
	
	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {br = new BufferedReader(new InputStreamReader(system));}
		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public String nextLine()throws IOException{return br.readLine();}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		public double nextDouble() throws IOException {return Double.parseDouble(next());}
		public char nextChar()throws IOException{return next().charAt(0);}
		public Long nextLong()throws IOException{return Long.parseLong(next());}
		public boolean ready() throws IOException{return br.ready();}
		public void waitForInput(){for(long i = 0; i < 3e9; i++);}
	}
}