CF651B-Beautiful Paintings

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.

一道如此水的题，不造怎么做了这么久，，真是无语，，学长说和又见拦截导弹差不多，于是我回去看了看，结果并没有什么收获，，但学长和我讲的时候倒是给了我点启发。。之前一直卡在元素重复问题上不造怎么解决，，原来那么水，，还是思路不行~~~

代码详解：

#include<bits/stdc++.h>
using namespace std;
const int N=1000+10;
int a[N];
int main()
{
    int n,i,j,x;
    while(~scanf("%d",&n))
    {
        x=0;
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        int k=1;
        for(i=0;i<n-1;i++)
        {
            for(j=k;j<n;j++)
            if(a[j]>a[i])
        {
            k=j+1;//下次直接从k开始找，避免重复问题；
            x++;
            break;
        }
        }
        printf("%d\n",x);
    }
    return 0;
}