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 long long int inf = 1e15;
const long long int mod = 1000000007;
int a[5010], n;
int st[20002], ans[5010], A[5010];
void construcST(int l = 1, int r = n, int i = 0) {
  if (l == r) {
    st[i] = l;
    return;
  }
  int mid = (l + r) / 2;
  construcST(l, mid, i * 2 + 1);
  construcST(mid + 1, r, i * 2 + 2);
  st[i] =
      (a[st[i * 2 + 1] >= a[st[i * 2 + 2]]]) ? st[i * 2 + 1] : st[i * 2 + 2];
}
int query(int s, int e, int l = 1, int r = n, int i = 0) {
  if (s > r or l > e) return -1;
  if (s <= l and r <= e) return st[i];
  int mid = (l + r) / 2;
  int left = query(s, e, l, mid, i * 2 + 1);
  int right = query(s, e, mid + 1, r, i * 2 + 2);
  if (left == -1)
    st[i] = right;
  else if (right == -1)
    st[i] = left;
  else
    st[i] = (a[left] >= a[right]) ? left : right;
  return st[i];
}
void update(int pos, int l = 1, int r = n, int i = 0) {
  if (pos < l or pos > r) return;
  if (l == r) {
    a[l]++;
    st[i] = l;
    return;
  }
  int mid = (l + r) / 2;
  update(pos, l, mid, i * 2 + 1);
  update(pos, mid + 1, r, i * 2 + 2);
  st[i] =
      (a[st[i * 2 + 1]] >= a[st[i * 2 + 2]]) ? st[i * 2 + 1] : st[i * 2 + 2];
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
  }
  for (int i = 0; i < n; i++) {
    memset(st, 0, sizeof st);
    memset(a, 0, sizeof a);
    for (int j = i; j < n; j++) {
      update(A[j]);
      ans[query(1, n)]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}
