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;
long long int apSum(long long int a, long long int n, long long int d = 1) {
  long long int res = (2 * a + (n - 1) * d) * n;
  res /= 2;
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isPrime(long long int n) {
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void solve() {
  long long int n;
  cin >> n;
  long long int list[n];
  for (long long int i = 0; i < n; i++) {
    cin >> list[i];
  };
  vector<long long int> arr(n + 1, 0);
  for (long long int i = 0; i < n; i++) {
    vector<long long int> local(n + 1, 0);
    long long int previos = -1;
    for (long long int j = i; j < n; j++) {
      long long int now = list[j], ans;
      local[now] += 1;
      if (previos == -1) {
        ans = now;
      } else {
        if (local[previos] < local[now])
          ans = now;
        else if (local[previos] > local[now])
          ans = previos;
        else if (now < previos)
          ans = now;
        else
          ans = previos;
      }
      arr[ans] += 1;
      previos = ans;
    }
  }
  for (long long int i = 1; i < n + 1; i++) {
    cout << arr[i] << " ";
  };
  cout << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int testCases = 1;
  while (testCases--) {
    solve();
  }
}
