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;
struct __s {
  __s() {
    srand(time(NULL));
    if (1) {
      ios_base::Init i;
      cin.sync_with_stdio(0);
      cin.tie(0);
    }
  }
  ~__s() {
    if (!1)
      fprintf(stderr, "Execution time: %.3lf s.\n",
              (double)clock() / CLOCKS_PER_SEC);
    long long n;
    cin >> n;
  }
} __S;
long long t[5555];
long long a[5555];
int main(void) {
  long long n;
  cin >> n;
  for (long long i = 0; i < (long long)(n); i++) {
    cin >> t[i];
    t[i]--;
  }
  for (long long l = 0; l < n; l++) {
    vector<long long> v;
    priority_queue<pair<long long, long long> > q;
    for (long long i = 0; i < (long long)(n); i++) {
      v.push_back(0);
      q.push(make_pair(0, -i));
    }
    for (long long r = l; r < n; r++) {
      v[t[r]]++;
      q.push(make_pair(v[t[r]], -t[r]));
      while (true) {
        long long res = q.top().first;
        long long i = -q.top().second;
        if (v[i] != res)
          q.pop();
        else
          break;
      }
      long long i = -q.top().second;
      a[i]++;
    }
  }
  for (long long i = 0; i < (long long)(n); i++) {
    cout << a[i] << " ";
  }
  cout << '\n';
  return 0;
}
