Codeforces Round #345 (Div. 2)——B. Beautiful Paintings（贪心求上升序列个数）

B. Beautiful Paintings

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n pictures delivered for the new exhibition. The i-th painting has beauty ai. We know that a visitor becomes happy every time he passes from a painting to a more beautiful one.

We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of a in any order. What is the maximum possible number of indices i (1 ≤ i ≤ n - 1), such that ai + 1 > ai.

Input

The first line of the input contains integer n (1 ≤ n ≤ 1000) — the number of painting.

The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000), where ai means the beauty of the i-th painting.

Output

Print one integer — the maximum possible number of neighbouring pairs, such that ai + 1 > ai, after the optimal rearrangement.

Examples

input

5
20 30 10 50 40

output

4

input

4
200 100 100 200

output

2

Note

In the first sample, the optimal order is: 10, 20, 30, 40, 50.

In the second sample, the optimal order is: 100, 200, 100, 200.

题意：求最多的上升序列的个数，用map比数组快多，做法：切蛋糕一样每次切一层

代码：

#include<iostream>
#include<algorithm>
#include<vector>
#include<sstream>
#include<map>
using namespace std;
int main(void)
{
	int t,n,i,j,be,sum,ans;
	while (cin>>n)
	{
		map<int,int>list;
		for (i=0; i<n; i++)
		{
			cin>>be;
			list[be]++;
		}
		ans=0;
		while (n)//每次
		{
			int num=0;
			for (map<int,int>::iterator it=list.begin(); it!=list.end(); it++)
			{
				if(it->second>0)
				{
					(it->second)--;
					n--;
					num++;
				}
			}
			ans=ans+num-1;//两个以上才算一个上升序列
		}
		cout<<ans<<endl;
	}
	return 0;
}