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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
long long int arr[6000], brr[6000], crr[6000], ans[6000];
int main() {
  int test_case = 1;
  for (int cs = 1; cs <= test_case; cs++) {
    long long int n;
    scanf("%lld", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &arr[i]);
    }
    for (long long int i = 1; i <= n; i++) {
      long long int a = n + 1;
      long long int b = 0;
      long long int c = 0;
      for (int j = 1; j <= n; j++) brr[j] = 0, crr[j] = n + 1;
      for (long long int j = i; j <= n; j++) {
        brr[arr[j]]++;
        crr[arr[j]] = min(crr[arr[j]], j);
        if (brr[arr[j]] > b || (brr[arr[j]] == b && arr[j] < a))
          b = brr[arr[j]], a = arr[j];
        ans[a]++;
      }
    }
    for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
  }
  return 0;
}
