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;
template <class T>
T min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <class T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
int n, a[5005], cnt[5005], ans[5005];
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) scanf("%d", &a[i]);
  for (int l = (1); l <= (n); ++l) {
    int curMax = 0, curCol = 0;
    memset(cnt, 0, sizeof cnt);
    for (int r = (l); r <= (n); ++r) {
      ++cnt[a[r]];
      if (cnt[a[r]] > curMax || (cnt[a[r]] == curMax && a[r] < curCol)) {
        curMax = cnt[a[r]];
        curCol = a[r];
      }
      ++ans[curCol];
    }
  }
  for (int i = (1); i <= (n); ++i) printf("%d ", ans[i]);
  return 0;
}
