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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
bool iseven(long long int n) {
  if ((n & 1) == 0) return true;
  return false;
}
void print(long long int t) { cout << t; }
void print(int t) { cout << t; }
void print(string t) { cout << t; }
void print(char t) { cout << t; }
void print(double t) { cout << t; }
void print(long double t) { cout << t; }
template <class T, class V>
void print(pair<T, V> p);
template <class T>
void print(vector<T> v);
template <class T>
void print(set<T> v);
template <class T, class V>
void print(map<T, V> v);
template <class T>
void print(multiset<T> v);
template <class T, class V>
void print(pair<T, V> p) {
  cout << "{";
  print(p.first);
  cout << ",";
  print(p.second);
  cout << "}";
}
template <class T>
void print(vector<T> v) {
  cout << "[ ";
  for (T i : v) {
    print(i);
    cout << " ";
  }
  cout << "]";
}
template <class T>
void print(set<T> v) {
  cout << "[ ";
  for (T i : v) {
    print(i);
    cout << " ";
  }
  cout << "]";
}
template <class T>
void print(multiset<T> v) {
  cout << "[ ";
  for (T i : v) {
    print(i);
    cout << " ";
  }
  cout << "]";
}
template <class T, class V>
void print(map<T, V> v) {
  cout << "[ ";
  for (auto i : v) {
    print(i);
    cout << " ";
  }
  cout << "]";
}
long long int freq[50005];
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n + 1);
  vector<long long int> ans(n + 1, 0);
  for (long long int i = 1; i <= n; ++i) cin >> a[i];
  for (long long int i = 1; i <= n; ++i) {
    for (long long int k = 1; k <= n; k++) freq[k] = 0;
    long long int currdom = 0;
    for (long long int j = i; j <= n; ++j) {
      ++freq[a[j]];
      if (freq[a[j]] > freq[currdom] or
          (freq[a[j]] == freq[currdom] and a[j] < currdom))
        currdom = a[j];
      ++ans[currdom];
    }
  }
  for (long long int i = 1; i <= n; ++i) cout << ans[i] << " ";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
