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;
int color[5010][5010];
int dom[5010][5010];
int cnt[5010];
int n;
void solve() {
  scanf("%d", &n);
  for (int(i) = 0; (i) < (int)(n); (i)++) {
    int c;
    scanf("%d", &c);
    c--;
    for (int(j) = 0; (j) < (int)(n); (j)++) color[j][i + 1] = color[j][i];
    color[c][i + 1]++;
    dom[i][i] = c;
    cnt[c]++;
  }
  for (int st = 0; st < n; st++) {
    for (int ed = st + 1; ed < n; ed++) {
      pair<int, int> cand1 = {
          color[dom[st][ed - 1]][ed + 1] - color[dom[st][ed - 1]][st],
          -dom[st][ed - 1]};
      pair<int, int> cand2 = {
          color[dom[ed][ed]][ed + 1] - color[dom[ed][ed]][st], -dom[ed][ed]};
      dom[st][ed] = -max(cand1, cand2).second;
      cnt[dom[st][ed]]++;
    }
  }
  for (int(i) = 0; (i) < (int)(n); (i)++) printf("%d ", cnt[i]);
  printf("\n");
}
int main() { solve(); }
