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 b[5000];
int ans[5001];
int sum[5001];
signed main() {
  int a;
  cin >> a;
  for (int c = 0; c < a; c++) cin >> b[c];
  for (int c = 0; c < a; c++) {
    memset(sum, 0, sizeof(sum));
    int MAX = 0;
    for (int d = c; d < a; d++) {
      sum[b[d]]++;
      if (sum[MAX] > sum[b[d]]) {
        ans[MAX]++;
      } else if (sum[MAX] < sum[b[d]]) {
        ans[b[d]]++;
        MAX = b[d];
      } else {
        if (MAX <= b[d]) {
          ans[MAX]++;
        } else {
          ans[b[d]]++;
          MAX = b[d];
        }
      }
    }
  }
  for (int i = 0; i < a; i++) {
    if (i) cout << " ";
    cout << ans[i + 1];
  }
  cout << endl;
}
