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;
const long long MOD = 1e9 + 7;
int main() {
  long long x, n, k, p;
  cin >> x;
  vector<long long> s(x + 1, 0);
  long long best = 0;
  vector<long long> ans(x + 1, 0);
  for (n = 1; n <= x; n++) cin >> s[n];
  for (n = 1; n <= x; n++) {
    vector<long long> cnt(x + 1, 0);
    for (k = n; k <= x; k++) {
      p = s[k];
      cnt[p]++;
      if (cnt[p] > cnt[best])
        best = p;
      else if (cnt[p] == cnt[best] && best > p)
        best = p;
      ans[best]++;
    }
  }
  for (n = 1; n <= x; n++) cout << ans[n] << ' ';
}
