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.
#include <bits/stdc++.h>
using namespace std;
int c[5001][5001];
int co[5001];
int x[5001];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) c[0][i] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> x[i];
    for (int j = 1; j <= n; j++) {
      if (x[i] == j)
        c[i][j] = c[i - 1][j] + 1;
      else
        c[i][j] = c[i - 1][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    co[x[i]] = co[x[i]] + 1;
    int d = 1;
    int f = x[i];
    for (int j = i + 1; j <= n; j++) {
      int h = c[j][x[j]] - c[i - 1][x[j]];
      if (x[j] != f) {
        if (h > d) {
          co[x[j]] = co[x[j]] + 1;
          f = x[j];
          d = h;
        } else {
          if (h == d) {
            if (f > x[j]) {
              co[x[j]] = co[x[j]] + 1;
              f = x[j];
              d = h;
            } else {
              co[f] = co[f] + 1;
            }
          } else
            co[f]++;
        }
      } else {
        co[f] = co[f] + 1;
        d++;
      }
    }
  }
  for (int i = 1; i <= n; i++) cout << co[i] << " ";
  cout << endl;
}
