C. Reorder the Array

You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.

For instance, if we are given an array [10,20,30,40][10,20,30,40], we can permute it so that it becomes [20,40,10,30][20,40,10,30]. Then on the first and the second positions the integers became larger (20>1020>10, 40>2040>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 22. Read the note for the first example, there is one more demonstrative test case.

Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.

Input

The first line contains a single integer nn (1≤n≤1051≤n≤105) — the length of the array.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the elements of the array.

Output

Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.

Examples

input

Copy

7
10 1 1 1 5 5 3

output

Copy

4

input

Copy

5
1 1 1 1 1

output

Copy

0

题意：给你一串数字，打乱顺序之后，每个位置上的数字如果比原来大的话，结果+1.

在进行排序之后优先使用最小的数字进行对比即可。

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#include<stack>
#include<cmath>
#include<set>
#include<map>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long

int a[100005];
int main()
{
	int n,i,j=0;
	cin>>n;
	for(i=0;i<n;i++)
	cin>>a[i];
	sort(a,a+n);
	for(i=1;i<n;i++)
	{
		if(a[i]>a[j])
		j++;
	}
	cout<<j<<endl;
	return 0;
}