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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.Scanner;
import java.io.PrintWriter;

public class TaskC {

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    public void solve(int testNumber, Scanner in, PrintWriter out) {

        int n = in.nextInt();

        int[] arr = new int[n];
        for(int i = 0 ; i < n ; i ++) {
            arr[i] = in.nextInt() - 1;
        }

        int[] rs = new int[n];

        int[] cnt = new int[n];
        int best = -1;
        double a = 0;
        for(int i = 0 ; i < n ; i ++) {
            Arrays.fill(cnt, 0);
            best = -1;
            for(int j = i ; j < n ;j ++) {
                cnt[arr[j]] += 1;
                a = Math.random();
                a = Math.random();
                if( best == -1) {
                    best = arr[j];
                }
                else {
                    if(cnt[best] < cnt[arr[j]]) {
                        best =  arr[j];
                    } else if (cnt[best] == cnt[arr[j]]){
                        if(best < arr[j]){
                            // do nothing
                        } else {
                            best = arr[j];
                        }
                    } else {
                        // do nothing
                    }
                }
                rs[best] += 1;
            }
        }
        for(int i = 0 ; i < n ; i ++) {
            out.print(rs[i] + " ");
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

    }
}
