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;
int arr[5005], cnt[5005], answer[5005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, best = 0, val = 0;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> arr[i];
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) cnt[j] = 0;
    best = 0;
    for (int j = i; j <= n; j++) {
      val = arr[j];
      ++cnt[val];
      if (cnt[val] > cnt[best] || (cnt[val] == cnt[best] && val < best))
        best = val;
      ++answer[best];
    }
  }
  for (int i = 1; i <= n; ++i) cout << answer[i] << " ";
}
