codeforces 651B

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.

第一行输入画的总数n，第二行输入每个画的“质量”，求出n个数中递增的数的个数。

我的想法是先找出最大的递增数列，然后将剩下的数组成一个数组在进行运算，直至数组个数为1

#include<iostream>
#include<algorithm>
#define MAXN 1002
using namespace std;
int increase(int *p, int len)
{
    int paint[MAXN], n=0, num=0;
    if (len <= 1)
        return 0;
    else
    {
        for (int i = 1;i < len;i++)
        {
            if (p[i] > p[i-1])
                num++;
            else
            {
                paint[n] = p[i];
                n++;
            }
        }
        return num + increase(paint, n);
    }
}
int main()
{
    int n, paint[MAXN], num;
    cin >> n;
    for (int i = 0;i < n;i++)
    {
        cin >> paint[i];
    }
    sort(paint, paint + n);
    num = increase(paint, n);
    cout << num << endl;
    return 0;
}