codeforces #345 div.2 B 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 a_i. 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 a_i + 1 > a_i.

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, ..., a_n (1 ≤ a_i ≤ 1000), where a_i means the beauty of the i-th painting.

Output

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

Examples

input

520 30 10 50 40

output

4

input

4200 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.

仔细观察后易发现，对于一个不含重复元素的集合来说，产生的最大幸福度是集合元素个数-1；那么只要把输入元素划分为数个不含相同元素的集合，分别求解即可；

#include<bits/stdc++.h>using namespace std;int main(){    int n;    scanf("%d",&n);    bool s[1001][1001]={0};    int num[1001]={0};    int col=0;    int temp;    for(int i=0;i<n;++i)    {        scanf("%d",&temp);        for(int j=1;j<=1000;++j)        {            if(s[j][temp]==0)            {                s[j][temp]=1;                ++num[j];                if(j>col)                    ++col;                break;            }        }    }    int ans=0;    for(int i=1;i<=col;++i)    {        ans+=num[i]-1;    }    printf("%d\n",ans);

    return 0;}