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 mod = 1e9 + 7;
const int maxn = 5001;
int a[maxn], cnt[maxn], ans[maxn];
pair<int, int> dp[maxn][maxn];
int main() {
  int i, n, j, u, v, m1, m2;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    ans[a[i]]++;
  }
  for (i = 1; i <= n; i++) {
    dp[i][i].second = 1;
    dp[i][i].first = a[i];
  }
  for (i = 1; i <= n; i++) {
    memset(cnt, 0, sizeof(cnt));
    cnt[a[i]]++;
    for (j = i + 1; j <= n; j++) {
      cnt[a[j]]++;
      if (cnt[a[j]] == dp[i][j - 1].second) {
        if (a[j] < dp[i][j - 1].first) {
          dp[i][j].second = cnt[a[j]];
          dp[i][j].first = a[j];
          ans[a[j]]++;
        } else {
          dp[i][j].second = dp[i][j - 1].second;
          ans[dp[i][j - 1].first]++;
          dp[i][j].first = dp[i][j - 1].first;
        }
      } else if (cnt[a[j]] > dp[i][j - 1].second) {
        dp[i][j].second = cnt[a[j]];
        dp[i][j].first = a[j];
        ans[a[j]]++;
      } else {
        dp[i][j].second = dp[i][j - 1].second;
        ans[dp[i][j - 1].first]++;
        dp[i][j].first = dp[i][j - 1].first;
      }
    }
  }
  for (i = 1; i <= n; i++) cout << ans[i] << " ";
}
