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 int maxN = 5000;
int colour[maxN + 5], ans[maxN + 5];
void read(int n) {
  for (int i = 1; i <= n; i++) cin >> colour[i];
}
void print(int n) {
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << "\n";
}
void dominant(int start, int end) {
  int index[maxN + 5];
  int count = 0, id = colour[start];
  memset(index, 0, sizeof(index));
  for (int i = start; i <= end; i++) {
    index[colour[i]]++;
    if (index[colour[i]] > count ||
        (index[colour[i]] == count && colour[i] < id)) {
      count = index[colour[i]];
      id = colour[i];
    }
    ans[id]++;
  }
}
int main() {
  int n;
  cin >> n;
  memset(colour, 0, sizeof(colour));
  memset(ans, 0, sizeof(ans));
  read(n);
  for (int i = 1; i <= n; i++) {
    dominant(i, n);
  }
  print(n);
  return 0;
}
