CodeForces 673C

Submit Status Practice CodeForces 673C

Description

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 t₁, t₂, ..., 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.

Sample Input

Input

41 2 1 2

Output

7 3 0 0 

Input

31 1 1

Output

6 0 0 

Hint

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 <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int N = 5050;
int a[N], num[N], v[N];


int main()
{
    int n;
    while(scanf("%d", &n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d", &a[i]);
        }
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;i++)
        {
            int sum1=0,sum2=0;
            memset(v,0,sizeof(v));
            for(int j=i;j<=n;j++)
            {
                v[a[j]]++;
                int h=v[a[j]];
                if(sum1<h||(sum1==h&&sum2>a[j]))
                {
                    sum1=h;
                    sum2=a[j];
                    num[a[j]]++;
                }
                else
                {
                    num[sum2]++;
                }
            }
        }
        for(int i=1;i<=n;i++)
        {
            printf("%d%c",num[i],i==n?'\n':' ');
        }
    }
    return 0;
}