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>
#pragma comment(linker, "/STACK:536870912")
const int MOD = 1000000007;
const int INF = 1000000001;
const int MAXN = 100010;
const long double EPS = 1e-6;
const int HASH_POW = 29;
const long double PI = acos(-1.0);
using namespace std;
void my_return(int code) { exit(code); }
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  mt19937 mt_rand(time(0));
  int n, p[5010], cnt[5010], ans[5010];
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i]);
  memset(ans, 0, sizeof ans);
  for (int l = 1; l <= n; ++l) {
    memset(cnt, 0, sizeof cnt);
    int mx = 0, pos = 1;
    for (int r = l; r <= n; ++r) {
      ++cnt[p[r]];
      if (cnt[p[r]] > mx || (cnt[p[r]] == mx && p[r] < pos)) {
        mx = cnt[p[r]];
        pos = p[r];
      }
      ++ans[pos];
    }
  }
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  my_return(0);
}
