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.Scanner;
import java.util.TreeMap;

public class Main {
  static int n;
  static int[] a;

  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);

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

    int[] ans = new int[n];
    for(int s = 0; s < n; s++){
      int[] counts = new int[n];
      int max = 1, maxi = a[s];
      for(int t = s; t < n; t++){
        ++counts[a[t]];
        int count = counts[a[t]];
        if(count >= max){
          if(count == max){
            if(maxi > a[t]){
              maxi = a[t];
            }
          }
          else{
            maxi = a[t];
            max = count;
          }
        }
        ++ans[maxi];
      }
    }

    for(int i = 0; i < n; i++){
      System.out.print(ans[i] + ((i == n -1) ? "\n" : " "));
    }
  }
}
