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 double EPS = 1e-8;
const double Pi = acos(-1);
bool inline equ(double a, double b) { return fabs(a - b) < EPS; }
int _R(int &x) { return scanf("%d", &x); }
int _R(long long &x) { return scanf("%" PRId64, &x); }
int _R(double &x) { return scanf("%lf", &x); }
int _R(char *s) { return scanf("%s", s); }
int R() { return 0; }
template <typename T, typename... U>
int R(T &x, U &...tail) {
  return _R(x) + R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%" PRId64, x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
const int N = 5010;
int n;
int ip[N];
int cnt[N];
int ans[N];
int main() {
  R(n);
  for (int i = (1); i <= (n); i++) R(ip[i]);
  for (int i = (1); i <= (n); i++) {
    int cmx = 0, cid = 0;
    fill(cnt, cnt + n + 1, 0);
    for (int j = (i); j <= (n); j++) {
      cnt[ip[j]]++;
      if (cnt[ip[j]] > cnt[cid] or (cnt[ip[j]] == cnt[cid] and ip[j] < cid)) {
        cid = ip[j];
      }
      ans[cid]++;
    }
  }
  for (int i = (1); i <= (n); i++) printf("%d%c", ans[i], " \n"[i == n]);
}
