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 n, m;
int a[100000];
set<long long> s;
long long ans[100000];
long long p[100000];
long long m1, m2;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 0; i < n; i++) {
    for (int i = 0; i < n; i++) {
      p[i] = 0;
    }
    for (int o = i; o < n; o++) {
      p[a[o]]++;
      if (p[a[o]] > m1) {
        m1 = p[a[o]];
        m2 = a[o];
      }
      if (p[a[o]] == m1 && m2 > a[o]) {
        m1 = p[a[o]];
        m2 = a[o];
      }
      ans[m2]++;
    }
    m1 = 0;
    m2 = 0;
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] << ' ';
  }
}
