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 comment(linker, "/STACK:216000000")
using namespace std;
const long long MAX = numeric_limits<long long>::max();
const long long MIN = numeric_limits<long long>::min();
const double PI = 3.1415926535;
const long long MOD = 1000000007;
template <class T>
ostream& operator<<(ostream& out, vector<T>& v) {
  for (int i = 0; i < v.size(); ++i) out << v[i] << " ";
  out << endl;
  return out;
}
template <class T>
istream& operator>>(istream& in, vector<T>& v) {
  for (int i = 0; i < v.size(); ++i) in >> v[i];
  return in;
}
template <class R, class L>
istream& operator>>(istream& in, pair<R, L>& p) {
  in >> p.first >> p.second;
  return in;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long ceil(long long a, long long b) {
  if (a % b) return a / b + 1;
  return a / b;
}
long long mod(long long a, long long b) { return (a % b + b) % b; }
long long inv_by_mod(long long x, long long m) {
  if (x == 0) return 1;
  long long ans = 1;
  long long cur = x;
  long long pow = m - 2;
  while (pow > 0) {
    if (pow & 1) ans = mod(ans * cur, m);
    pow >>= 1;
    cur = mod(cur * cur, m);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  cin >> a;
  for (int i = 0; i < n; ++i) {
    a[i]--;
  }
  vector<int> count(n);
  for (int i = 0; i < n; ++i) {
    vector<int> ccur(n);
    int dom = a[i];
    for (int j = i; j < n; ++j) {
      ccur[a[j]]++;
      if (ccur[a[j]] > ccur[dom])
        dom = a[j];
      else if (ccur[a[j]] == ccur[dom]) {
        dom = min(dom, a[j]);
      }
      count[dom]++;
    }
  }
  cout << count;
}
