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;
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long nCr(long long n, long long r) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = min(i, r); j > 0; j--) C[j] = C[j] + C[j - 1];
  return C[r];
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n == 2) return true;
  if (n % 2 == 0) return false;
  long long m = sqrt(n);
  for (long long i = 3; i <= m; i += 2)
    if (n % i == 0) return false;
  return true;
}
long long M(long long a) { return (a % (1000000007LL)); }
bool isPowerOfTwo(long long x) { return x && (!(x & (x - 1))); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> v(n);
  for (long long i = 0; i < (long long)(n); ++i) cin >> v[i];
  vector<int> ans(n + 1);
  for (long long i = 0; i < (long long)(n); ++i) {
    vector<int> color_total(n + 1);
    int best = 0;
    for (int j = i; j < n; j++) {
      color_total[v[j]]++;
      if (color_total[v[j]] > color_total[best]) {
        best = v[j];
      } else if (color_total[v[j]] == color_total[best]) {
        if (v[j] < best) {
          best = v[j];
        }
      }
      ans[best]++;
    }
  }
  for (long long i = 1; i <= (long long)(n); ++i) {
    cout << ans[i] << " ";
  }
  return 0;
}
