CodeForces 651B Beautiful Paintings

http://codeforces.com/problemset/problem/651/B

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：

给定一些数字，问最多组成多少对右边大于左边数字的序列。

思路：

头脑简单的用了下 sort 结果不对昂。其实，

只要记录出现最多的数字个数就行了，然后用总个数减去该数字个数就过了。

AC CODE：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define retrun return
#define mian main
#define ture true
using namespace std;
const int MYDD=1103;

int main() {
	int n,a[MYDD];
	scanf("%d",&n);
	memset(a,0,sizeof(a));
	int Appear=-1;/*记录出现次数最多的数字个数*/
	for(int j=0; j<n; j++) {
		int t;
		scanf("%d",&t);
		a[t]++;
		if(a[t]>Appear)  Appear=a[t];
	}

	int ans=n-Appear;
	printf("%d\n",ans);
	return 0;
}