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 v[5005], frecv[5005];
void fs(int k, int n) {
  int x[5005], max = 0, poz;
  memset(x, 0, sizeof(x));
  x[v[k]]++;
  x[v[++k]]++;
  if (v[k] == v[k - 1]) {
    max = 2;
    poz = v[k];
  }
  if (v[k] < v[k - 1]) {
    max = 1;
    poz = v[k];
  }
  if (v[k] > v[k - 1]) {
    max = 1;
    poz = v[k - 1];
  }
  frecv[poz]++;
  while (k < n) {
    k++;
    if (x[v[k]] == 0) {
      x[v[k]] = 1;
      if (max == 1 && v[k] < poz) poz = v[k];
      frecv[poz]++;
      continue;
    } else {
      x[v[k]]++;
      if (x[v[k]] > max) {
        max = x[v[k]];
        poz = v[k];
      }
      if (x[v[k]] == max && v[k] < poz) poz = v[k];
      frecv[poz]++;
    }
  }
}
int main() {
  int n, i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
    frecv[v[i]]++;
  }
  for (int k = 1; k <= n - 1; k++) fs(k, n);
  for (i = 1; i <= n; i++) printf("%d ", frecv[i]);
  return 0;
}
