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.*;
import java.util.*;
import static java.util.Collections.*;
import static java.lang.Math.*;

@SuppressWarnings("unchecked")
public class A_Bear_and_Colors {
    public static PrintWriter out;
    public static InputReader in;
    //lul
    public static void main(String[] args)throws IOException {
        in = new InputReader(System.in);
        out = new PrintWriter(System.out);
        
        int cases = 1;
        
        for(int t = 0; t < cases; t++){
            long n = in.nextLong();
            long arr[] = new long[(int)n];
            long ct[] = new long[(int)n];
            int pos[] = new int[(int)n]; Arrays.fill(pos,(int)n+1);
            long ans[] = new long[(int)n]; Arrays.fill(ans,0l);
            for(int i=0;i<n;i++) {arr[i] = in.nextLong()-1; pos[(int)arr[i]] = min(i,pos[(int)arr[i]]);}
            for(int i=0;i<n;i++){
                long mx = -1;
                Arrays.fill(ct,0l);
                // Arrays.fill(pos,(int)n+1);
                for(int j=i;j<n;j++){
                    ct[(int)arr[j]]++;
                    long cnt = ct[(int)arr[j]];
                    // pos[(int)arr[j]] = min(j,pos[(int)arr[j]]);
                    if(mx==-1 || cnt > ct[(int)mx] || (arr[j]<mx && cnt==ct[(int)mx])){
                        mx = arr[j];
                    }
                    // out.println("i="+i+",j="+j+",mx="+mx);
                    ans[(int)mx]++;
                }
            }
            for(int i=0;i<n;i++) out.print(ans[i]+" "); out.println();
        }

        out.close();
    }
    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());
        }

    }

}