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;
const long long nax = 1e5 + 5;
vector<long long> adj[nax];
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 7;
long long m_ans[nax];
long long cnt[nax];
long long a[nax];
signed main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long majority = 0;
  for (long long i = 0; i < n; i++) {
    memset(cnt, 0, sizeof(cnt));
    majority = a[i];
    cnt[majority] = 1;
    m_ans[majority]++;
    ;
    for (long long j = i + 1; j < n; j++) {
      cnt[a[j]]++;
      if (cnt[majority] == cnt[a[j]]) {
        if (majority < a[j]) {
          m_ans[majority]++;
          continue;
        } else {
          m_ans[a[j]]++;
          majority = a[j];
          continue;
        }
      } else if (cnt[a[j]] > cnt[majority]) {
        majority = a[j];
        m_ans[majority]++;
        continue;
      } else {
        m_ans[majority]++;
        continue;
      }
    }
  }
  for (long long i = 1; i <= n; i++) {
    cout << m_ans[i] << "\n";
  }
  return 0;
}
