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;
std::vector<int> v;
pair<int, int> m[2][5001];
int cdad[5001][5001];
int res[5001];
int main() {
  int n, tmp;
  cin >> n;
  v.reserve(n);
  for (int i = 0; i < n; ++i) {
    cin >> tmp;
    v.push_back(tmp);
    m[0][i] = make_pair(1, tmp);
    res[tmp]++;
    cdad[i][tmp]++;
  }
  unsigned char z = 1;
  for (int j = 1; j < n; ++j) {
    for (int h = 0; h + j < n; ++h) {
      pair<int, int> sol = m[z ^ 1][h];
      int c = v[h + j];
      cdad[h][c]++;
      int x = cdad[h][c];
      if ((x > sol.first) or (x == sol.first and c < sol.second)) {
        sol.first = x;
        sol.second = c;
      }
      m[z][h] = sol;
      res[sol.second]++;
    }
    z ^= 1;
  }
  for (int i = 1; i <= n; ++i) {
    cout << res[i] << " ";
  }
  cout << endl;
}
