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>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
int a, b, c, d, n, m, k;
int mas[5002], cur[5002], ans[5002];
int main() {
  scanf("%d", &n);
  for (int _n((n)-1), i(0); i <= _n; i++) {
    scanf("%d", &mas[i]);
  }
  for (int _n((n)-1), i(0); i <= _n; i++) {
    priority_queue<pair<int, int> > q;
    memset((cur), 0, sizeof(cur));
    for (int _n(n - 1), j(i); j <= _n; j++) {
      ++cur[mas[j]];
      q.push(make_pair(cur[mas[j]], -mas[j]));
      while (!q.empty()) {
        auto t = q.top();
        if (cur[-t.second] != t.first) {
          q.pop();
        } else {
          break;
        }
      }
      ++ans[-q.top().second];
    }
  }
  for (int _n((n)), i(1); i <= _n; i++) {
    if (i > 1) printf(" ");
    printf("%d", ans[i]);
  }
  printf("\n");
}
