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;
template <class T>
T gcd(T a, T b) {
  return (b == 0) ? a : gcd(b, a % b);
}
const unsigned int MOD = 1000000007;
long long int inf = 1000000000000000000;
long long int MODI = 998244353;
template <class T>
T power(T a, long long int n) {
  T res = 1;
  while (n > 0) {
    if (n % 2) res = (res * a);
    a = (a * a), n = n / 2;
  }
  return res;
}
bool isprime(int n) {
  if (n <= 1) return false;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
bool issq(long long int n) {
  double d = n;
  double p1 = sqrt(d);
  long long int p2 = sqrt(n);
  return (p1 == p2);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  vector<long long int> arr;
  ;
  for (long long int i = 0; i <= n - 1; i++) {
    long long int a;
    cin >> a;
    arr.push_back(a);
  }
  map<long long int, long long int> ans;
  for (int i = 0; i <= n - 1; i++) {
    long long int maxx = 0, maxxind = 0;
    vector<long long int> maps(n + 1, 0);
    ;
    for (int j = i; j <= n - 1; j++) {
      maps[arr[j]]++;
      if (maps[arr[j]] > maxx) {
        maxx = maps[arr[j]];
        maxxind = arr[j];
        ans[arr[j]]++;
      } else if (maps[arr[j]] == maxx) {
        if (arr[j] < maxxind)
          ans[arr[j]]++, maxxind = arr[j];
        else
          ans[maxxind]++;
      } else
        ans[maxxind]++;
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
}
