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 N;
int A[5003];
int pre[5003][5003];
int dp[5003][5003];
int ans[5003];
int main(int argc, char** argv) {
  std::ios::sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> A[i];
  for (int k = 1; k <= N; k++) {
    for (int i = 1; i <= N; i++) {
      pre[k][i] = pre[k][i - 1] + (A[i] == k);
    }
  }
  for (int i = 1; i <= N; i++) {
    dp[i][i] = A[i];
    ans[A[i]]++;
  }
  for (int l = 2; l <= N; l++) {
    for (int i = 1; i + l - 1 <= N; i++) {
      int j = i + l - 1;
      int a1 = dp[i][j - 1];
      int a2 = A[j];
      int c1 = pre[a1][j] - pre[a1][i - 1];
      int c2 = pre[a2][j] - pre[a2][i - 1];
      int mx;
      if (c1 > c2)
        mx = a1;
      else if (c1 < c2)
        mx = a2;
      else
        mx = min(a1, a2);
      ans[mx]++;
      dp[i][j] = mx;
    }
  }
  for (int i = 1; i <= N; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
