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;
pair<int, int> pmax(pair<int, int> a, pair<int, int> b) {
  if (a.first > b.first)
    return a;
  else if (b.first > a.first)
    return b;
  else {
    if (a.second < b.second) return a;
    return b;
  }
}
int n, p = 1 << 13;
vector<pair<int, int> > puu(1 << 14, {0, 0});
int t[5050];
void add(int k, int x) {
  k += p;
  puu[k].first += x;
  k /= 2;
  while (k) {
    puu[k] = pmax(puu[k * 2], puu[k * 2 + 1]);
    k /= 2;
  }
}
void setp(int k, int x) {
  k += p;
  puu[k].first = x;
  k /= 2;
  while (k) {
    puu[k] = pmax(puu[k * 2], puu[k * 2 + 1]);
    k /= 2;
  }
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> t[i];
    puu[p + i + 1].second = i + 1;
  }
  int ans[5050] = {0};
  for (int i = 0; i < n; i++) {
    int cc[5050] = {0};
    int b = 0;
    for (int j = i; j < n; j++) {
      cc[t[j]]++;
      if (cc[t[j]] > cc[b]) {
        b = t[j];
      } else if (cc[t[j]] == cc[b]) {
        if (t[j] < b) b = t[j];
      }
      ans[b]++;
    }
    for (int j = i; j < n; j++) {
      setp(t[i], 0);
    }
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i + 1] << " ";
  }
  cout << endl;
  return 0;
}
