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 mod = 1000000007;
vector<long long> vec[3000005];
long long fastexpo(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    b = b / 2;
    a = (a * a) % mod;
  }
  return ans % mod;
}
long long dp[1000005] = {0};
bool vis[1000005] = {0};
long long bits(long long x) {
  long long num = 0;
  while (x) {
    if (!(x & 1)) {
      num++;
    }
    x /= 2;
  }
  return num;
}
long long countDivisibles(long long A, long long B, long long M) {
  if (A % M == 0) return (B / M) - (A / M) + 1;
  return (B / M) - (A / M);
}
int32_t main() {
  long long n;
  cin >> n;
  long long dp[n + 1], arr[n];
  for (long long i = 0; i < n; i++) cin >> arr[i];
  long long ans[n][n];
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) dp[j + 1] = 0;
    long long max1 = arr[i];
    for (long long j = i; j < n; j++) {
      dp[arr[j]]++;
      if (dp[arr[j]] > dp[max1]) {
        max1 = arr[j];
      } else {
        if (dp[arr[j]] == dp[max1]) {
          max1 = min(arr[j], max1);
        }
      }
      ans[i][j] = max1;
    }
  }
  for (long long i = 0; i < n; i++) {
    dp[i + 1] = 0;
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = i; j < n; j++) {
      dp[ans[i][j]]++;
    }
  }
  for (long long i = 1; i <= n; i++) {
    cout << dp[i] << " ";
  }
  return 0;
}
