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.
n = int(raw_input())
balls = list(map(int, raw_input().split()))
domination = (n + 1) * [ 0 ]

class Color:
  def __init__(self, id):
    self.id = id
    self.heap_pos = None

colors = [ Color(color) for color in range(n + 1) ]

for start in range(n):
  for color in colors:
    color.count = 0
  max_count = 0
  best_color = colors[1]
  for i in range(start, n):
    color = colors[balls[i]]
    color.count += 1
    if color.count == max_count:
      if color.id < best_color.id:
        best_color = color
    elif color.count > max_count:
      max_count = color.count
      best_color = color
    domination[best_color.id] += 1
print(' '.join(map(str, domination[1:])))
