CodeForces

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.

题目大意：可重排原数组，使得ai+1>ai的组数最大，求最大组数

题目分析：数据很小，无脑n^2模拟即可，记个O(n)做法，n减去出现次数最多的数的出现次数就是答案

#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
int main()
{
    int n,m=0,x,i;
    int a[1010];
    while(cin>>n){
        memset(a,0,sizeof(a));
        for(i=0;i<n;i++){
            cin>>x;
            m=max(m,a[x]++);
        }
        cout<<n-m-1<<endl;
    }
    return 0;
}