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 rdtsc() {
  long long tmp;
  asm("rdtsc" : "=A"(tmp));
  return tmp;
}
inline int myrand() { return rand(); }
inline int rnd(int x) { return myrand() % x; }
const int inf = (int)1.01e9;
const long double eps = 1e-9;
const long double pi = acos((long double)-1.0);
void precalc() {}
const int maxn = (int)5e3 + 10;
int a[maxn];
int n;
int read() {
  if (scanf("%d", &n) < 1) {
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    --a[i];
  }
  return 1;
}
int cnt[maxn], ans[maxn];
void solve() {
  for (int i = 0; i < n; ++i) {
    ans[i] = 0;
  }
  for (int l = 0; l < n; ++l) {
    for (int i = 0; i < n; ++i) {
      cnt[i] = 0;
    }
    pair<int, int> mx(0, 0);
    for (int r = l; r < n; ++r) {
      cnt[a[r]] += 1;
      mx = max(mx, make_pair(cnt[a[r]], -a[r]));
      ans[-mx.second] += 1;
    }
  }
  for (int i = 0; i < n; ++i) {
    printf("%d%c", ans[i], " \n"[i == n - 1]);
  }
}
int main() {
  srand(rdtsc());
  precalc();
  while (1) {
    if (!read()) {
      break;
    }
    solve();
  }
  return 0;
}
