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 n, a[5001];
int sum[5001][5001], d[5001][5001], res[5001];
int vet, moi;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) sum[i][j] = sum[i - 1][j] + (a[i] == j);
  for (int i = 1; i < n; i++) {
    res[a[i]]++;
    vet = a[i];
    d[i][i] = 1;
    for (int j = i + 1; j <= n; j++) {
      moi = vet;
      d[i][j] = d[i][j - 1];
      if (d[i][j] < sum[j][a[j]] - sum[i - 1][a[j]]) {
        moi = a[j];
        d[i][j] = sum[j][a[j]] - sum[i - 1][a[j]];
      } else if (d[i][j] == sum[j][a[j]] - sum[i - 1][a[j]]) {
        if (moi > a[j]) {
          moi = a[j];
          d[i][j] = sum[j][a[j]] - sum[i - 1][a[j]];
        }
      }
      res[moi]++;
      vet = moi;
    }
  }
  res[a[n]]++;
  for (int i = 1; i <= n; i++) cout << res[i] << ' ';
}
