Codeforces Round #351 Div2 C 简单枚举

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探讨了在一系列带有颜色的球中，如何通过编程计算每种颜色在所有可能的连续子区间内作为主导颜色出现的次数。文章提供了一种通过逐个检查区间并更新主导颜色计数的方法。

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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 $\text{[math]}$ 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.

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.

题意：

每个球对应一种颜色，一共有n个球，所以有 $\text{[math]}$ 个区间，问每种颜色在多少个区间中是dominant？

dominant的意思是在这个区间中这种颜色数最多，如果有相同个数的颜色，编号最小的是dominant。

分析：

枚举 $\text{[math]}$ 个区间，对于每个区间统计那种颜色是dominant即可。

#include<bits/stdc++.h>
using namespace std;
const int N=5009;
int ans[N],num[N],a[N];
int main()
{
    int n;scanf("%d",&n);
    for(int i=0;i<n;i++)scanf("%d",&a[i]);
    int tot=0;
    for(int i=0;i<n;i++){
        memset(num,0,sizeof(num));
        int maxn=0,maxp=i;
        for(int j=i;j<n;j++){
            tot++;
            num[a[j]]++;
            if(num[a[j]]>maxn||(a[j]<maxp&&num[a[j]]==maxn)){
                maxn=num[a[j]];
                maxp=a[j];
                //ans[a[j]]++;
            }
            ans[maxp]++;
        }
    }
   // cout<<tot<<endl;
    for(int i=1;i<=n;i++)printf("%d ",ans[i]);
    return 0;
}