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 maxim(int v[6000], int n) {
  int i, poz = 1, m = v[1];
  for (i = 1; i <= n; i++)
    if (v[i] > m) {
      m = v[i];
      poz = i;
    }
  return poz;
}
int culori[6000], cnt[6000];
int main() {
  int n, i, m, v[6000], l = 0, k, start, stop, j;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> v[i];
    culori[v[i]]++;
  }
  start = 1;
  while (start < n) {
    stop = start + 1;
    for (i = 0; i <= n; i++) cnt[i] = 0;
    cnt[v[start]]++;
    cnt[v[stop]]++;
    if (v[start] != v[stop])
      m = 1;
    else
      m = 2;
    if (v[start] > v[stop])
      k = v[stop];
    else
      k = v[start];
    culori[k]++;
    while (stop < n) {
      stop++;
      cnt[v[stop]]++;
      if (cnt[v[stop]] > m) {
        culori[v[stop]]++;
        m = cnt[v[stop]];
        k = v[stop];
      } else if (cnt[v[stop]] == m) {
        if (v[stop] > k) culori[k]++;
        if (v[stop] <= k) {
          culori[v[stop]]++;
          k = v[stop];
        }
      } else if (cnt[v[stop]] < m)
        culori[k]++;
    }
    start++;
  }
  for (i = 1; i <= n; i++) cout << culori[i] << " ";
  return 0;
}
