【codeforces 651B - Beautiful Paintings 】 greedy(贪心)

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：：游客在欣赏画时，通常会因为当前这幅画比前一幅画美丽而获得一点幸福度。 给出n副画作的美丽值，现在要将他们排成一行。求出游客能获得的最大幸福值。

思路：两个优先队列维护（互相倒）

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int n;
int ans;
int a[10005];
priority_queue<int> q1,q2;
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{

scanf("%d",&a[i]);
q1.push(-a[i]);
}
int now;
while((!q1.empty())||(!q2.empty()))
{

if(!q1.empty()) now=q1.top(),q1.pop();
while(!q1.empty())
{
int nex=q1.top();
q1.pop();
if(nex<now)
{
ans++;
now=nex;
}
else q2.push(nex);
}
if(!q2.empty()) now=q2.top(),q2.pop();
while(!q2.empty())
{
int nex=q2.top();
q2.pop();
if(nex<now)
{
ans++;
now=nex;
}
else q1.push(nex);
}
}
printf("%d\n",ans);
}