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 long double pi = acos(-1.0);
const int intINF = (int)1e9;
template <typename T>
T gcd(T a, T b) {
  while (a > 0 && b > 0) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return (a + b);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <typename T>
bool prime(T n) {
  if (n <= 1) {
    return false;
  }
  T i = 2;
  while (i * i <= n) {
    if (n % i == 0) {
      return false;
    }
    ++i;
  }
  return true;
}
long long get_power(long long a, long long b, long long mod) {
  if (b == 0) {
    return 1 % mod;
  }
  long long ans = get_power(a, b / 2, mod);
  ans = (ans * ans) % mod;
  if (b % 2) {
    ans *= a;
  }
  ans %= mod;
  return ans;
}
template <typename T, typename U>
U power(T a, T b, U modulo) {
  long long A = (long long)a, B = (long long)b, C = (long long)modulo;
  return (U)get_power(A, B, C);
}
const int maxN = 5.1e3;
int ans[maxN], a[maxN], cnt[maxN];
void solveTask() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) cnt[j] = 0;
    ++ans[a[i]];
    cnt[a[i]] = 1;
    int opt = a[i];
    for (int j = i + 1; j <= n; ++j) {
      ++cnt[a[j]];
      if (cnt[a[j]] > cnt[opt] || (cnt[a[j]] == cnt[opt] && a[j] < opt))
        opt = a[j];
      ++ans[opt];
    }
  }
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  solveTask();
  return 0;
}
