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 int maxn = 6000;
int a[maxn];
int ans[maxn];
int cnt[maxn];
int first[maxn];
bool solve() {
  int n;
  if (scanf("%d", &n) != -1) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      a[i]--;
      ans[i] = 0;
    }
    for (int i = 0; i < n; i++) {
      memset(cnt, 0, sizeof cnt);
      memset(first, 0, sizeof first);
      int curbest = 0;
      for (int j = i; j < n; j++) {
        cnt[a[j]]++;
        if (cnt[a[j]] == 1) {
          first[a[j]] = a[j];
        }
        if (cnt[a[j]] > cnt[curbest] ||
            (cnt[a[j]] == cnt[curbest] && first[a[j]] < first[curbest])) {
          curbest = a[j];
        }
        ans[curbest]++;
      }
    }
    for (int i = 0; i < n; i++) {
      printf("%d%c", ans[i], " \n"[i + 1 == n]);
    }
    return true;
  }
  return false;
}
int main() {
  auto t = clock() / CLOCKS_PER_SEC;
  while (solve()) {
    auto nt = clock() / CLOCKS_PER_SEC;
    cerr << "TIME: " << (nt - t) << endl;
    t = nt;
  }
  return 0;
}
