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;
const int N = 5 * int(1e3) + 3;
int T[N];
int Ans[N];
int Col[N];
bool comp(pair<int, int> a, pair<int, int> b) {
  if (a.first > b.first)
    return true;
  else {
    if (a.first == b.first) {
      if (a.second <= b.second)
        return true;
      else
        return false;
    } else {
      return false;
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> T[i];
  }
  for (int j = 0; j < n; ++j) {
    pair<int, int> Best = make_pair(0, 0);
    for (int k = j; k < n; ++k) {
      Col[T[k]]++;
      pair<int, int> Check = make_pair(Col[T[k]], T[k]);
      if (comp(Check, Best)) Best = Check;
      Ans[Best.second]++;
    }
    for (int t = 0; t <= n; ++t) {
      Col[t] = 0;
    }
  }
  for (int u = 1; u <= n; ++u) {
    cout << Ans[u] << " ";
  }
  return 0;
}
