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 pref[5001][5001];
int dp[5001][5001];
int mx(int sum1, int sum2, int v1, int v2) {
  if (sum1 == sum2)
    return min(v1, v2);
  else {
    if (sum1 > sum2) return v1;
    return v2;
  }
}
int main() {
  memset(pref, 0, sizeof(pref));
  memset(dp, 0, sizeof(dp));
  int n;
  cin >> n;
  vector<int> a(n + 1);
  vector<int> ans(n + 1, 0);
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      if (a[j] == i)
        pref[i][j] = pref[i][j - 1] + 1;
      else
        pref[i][j] = pref[i][j - 1];
    }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      if (i == j)
        dp[i][j] = a[i];
      else {
        int prv = dp[i][j - 1];
        int prs = a[j];
        int val = mx(pref[prv][j] - pref[prv][i - 1],
                     pref[prs][j] - pref[prs][i - 1], prv, prs);
        dp[i][j] = val;
      }
      ans[dp[i][j]]++;
    }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
}
