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 INF = 1000000404;
const long long MOD = 1000000007ll;
const long double PI = acos(-1.0);
const long double EPS = 1e-9;
template <typename t1, typename t2>
inline void upmax(t1 &a, t2 b) {
  a = max(a, (t1)b);
}
template <typename t1, typename t2>
inline void upmin(t1 &a, t2 b) {
  a = min(a, (t1)b);
}
template <typename T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
inline T lcm(T a, T b) {
  return a * (b / gcd(a, b));
}
template <typename T>
inline T sqr(T a) {
  return a * a;
}
template <typename T>
inline bool pal(T &x) {
  int n = (int((x.size())));
  for (int i = 0; i < n / 2; i++) {
    if (x[i] != x[n - i - 1]) return 0;
  }
  return 1;
}
template <typename T>
inline void rev(T &x) {
  int n = (int((x.size())));
  for (int i = 0; i < n / 2; i++) swap(x[i], x[n - i - 1]);
}
int month[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
inline long long mp(long long a, long long b) { return (a << 31) + b; }
class compare {
 public:
  bool operator()(const int a, const int b) const { return 1; }
};
int SQ = 400;
int res[5555];
int a[5555];
int cnt[5555];
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i < n + 1; i++) cin >> a[i];
  int maxx = 0;
  int ind = 0;
  for (int i = 1; i < n + 1; i++) {
    fill(cnt, cnt + 5555, 0);
    maxx = 0;
    for (int j = i; j < n + 1; j++) {
      cnt[a[j]]++;
      if (cnt[a[j]] > maxx) {
        maxx = cnt[a[j]];
        ind = a[j];
      } else {
        if (maxx == cnt[a[j]]) {
          upmin(ind, a[j]);
        }
      }
      res[ind]++;
    }
  }
  for (int i = 1; i < n + 1; i++) cout << res[i] << ' ';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  };
  getchar();
  getchar();
  return 0;
}
