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:167772160")
using namespace std;
const int INF = 1e9 + 10;
const long long LINF = (long long)2e18 + 10;
const long double PI = acos(-1);
const long double eps = 1e-8;
const long double EPS = 1e-12;
const int N = 5e3 + 10;
int n, cols[N], balls[N], res[N];
void solve() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &balls[i]);
    --balls[i];
  }
  for (int b = 0; b < n; ++b) {
    fill(cols, cols + N, 0);
    int maxc = -1;
    int maxcnt = -1;
    for (int i = b; i < n; ++i) {
      int ball = balls[i];
      ++cols[ball];
      if (cols[ball] > maxcnt || (cols[ball] == maxcnt && ball < maxc)) {
        maxcnt = cols[ball];
        maxc = ball;
      }
      ++res[maxc];
    }
  }
  for (int i = 0; i < n; ++i) printf("%d ", res[i]);
}
int main() {
  solve();
  return 0;
}
