codeforces 651B 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.

算是一个贪心吧，看数列的哪一个排列能够使相邻的两个元素前一个比后一个大的情况最多

先从大到小排序，然后用一个vis标记这个数有没有用过，找到第一个没有用过的最小的，然后往后扫尽可能多的连续的递增的数，反复这样做，直到所有的数都遍历一遍

需要注意的是有可能有重复的数字

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>

using namespace std;

int a[1005];
bool vis[1005];

int main()
{
	int n;
	memset(vis, false, sizeof(vis));
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	//升序排列
	sort(a, a + n);
	int cou = 0, ans = 0;

	while (cou != n) {
		int i;
		//找到第一个没有用过的最小的数a[i]
		for (i = 0; i < n; i++) {
			if (!vis[i]) {
				vis[i] = true;
				cou++;
				break;
			}
		}
		//从a[i + 1]开始找尽可能长的递增的数列
		for (int j = i + 1; j < n; j++) {
			if (!vis[j] && a[j] > a[i]) {
				vis[j] = true;
				cou++;
				ans++;
				i = j; // 记得改变i
			}
		}
	}
	printf("%d\n", ans);
	return 0;
}