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>
unsigned int C[5000 + 2][5000 + 2] = {{0}};
unsigned int D[5000 + 2][5000 + 2] = {{0}};
unsigned int R[5000 + 2] = {0};
static unsigned short count(int c, int i, int j) {
  return (C[j][c] - C[i - 1][c]);
}
int main(int argc, char *argv[]) {
  int n, t;
  int i, j;
  scanf("%d", &n);
  for (j = 1; j <= n && scanf("%d", &t) == 1; j++) {
    memcpy(&C[j][0], &C[j - 1][0], sizeof(C[0][0]) * (5000 + 1));
    C[j][t]++;
    for (i = j; i > 0; i--) {
      int c = D[j - 1][i], tij = count(t, i, j), cij = count(c, i, j);
      if (tij > cij || (tij == cij && t < c)) {
        D[j][i] = t;
      } else
        D[j][i] = c;
      R[D[j][i]]++;
    }
  }
  for (i = 1; i <= n; i++) {
    printf("%u%c", (unsigned int)R[i], (i == n) ? '\n' : ' ');
  }
  return 0;
}
