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 dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int mod = 1000000007;
const double pi = acos(-1);
const long long OO = INT_MAX;
double distance(double a, double b, double x, double y) {
  return sqrt((x - a) * (x - a) + ((y - b) * (y - b)));
}
int c[5000 + 1];
int Ans[5000 + 1];
int main() {
  cin.tie(0);
  std::ios::sync_with_stdio(false);
  int N;
  scanf("%d", &N);
  int A[N];
  for (int i = 0; i < N; i++) scanf("%d", &A[i]);
  for (int i = 0; i < N; i++) {
    memset(c, 0, sizeof(c));
    int mx = 0, dominant;
    for (int j = i; j < N; j++) {
      c[A[j]]++;
      if (c[A[j]] > mx || (c[A[j]] == mx && dominant > A[j])) {
        mx = c[A[j]];
        dominant = A[j];
      }
      Ans[dominant]++;
    }
  }
  for (int i = 0; i < N; i++) cout << Ans[i + 1] << ' ';
}
