Bear and Colors

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 t i.

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  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 t 1, t 2, …, t n (1 ≤ t i ≤ n) where t i 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.

Solution:

#include <bits/stdc++.h>

using namespace std;

const int N = 77777;

int cnt[N], a[N], ans[N];

int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    a[i]--;
  }
  for (int i = 0; i < n; i++) {
    ans[i] = 0;
  }
  for (int start = 0; start < n; start++) {
    for (int i = 0; i < n; i++) {
      cnt[i] = 0;
    }
    int best = 0;
    for (int i = start; i < n; i++) {
      cnt[a[i]]++;
      if (cnt[a[i]] > cnt[best] || (cnt[a[i]] == cnt[best] && a[i] < best)) {
        best = a[i];
      }
      ans[best]++;
    }
  }
  for (int i = 0; i < n; i++) {
    if (i > 0) {
putchar(' ');
}
printf("%d", ans[i]);
}
printf("\n");
return 0;
}