题目：

Reorder the Array

 CodeForces - 1007A 

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

7
10 1 1 1 5 5 3

Output

4

Input

5
1 1 1 1 1

Output

0

Note

In the first sample, one of the best permutations is [1,5,5,3,10,1,1][1,5,5,3,10,1,1]. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.

In the second sample, there is no way to increase any element with a permutation, so the answer is 0.

题意：

  给你一个序列，你可以生成这个序列的任意一个排列，对于某个排列，如果这个排列上某个位置的值大于原序列的值，那么就会产生1的贡献，问你最大能有多少贡献。新版田忌赛马！自己的马和自己的马比赛，最多赢几场。

思路：

  先进行排序，然数列呈从小到大分布，用 sort 。之后利用贪心思想，从大到小逐个判定。

题解：

#include <bits/stdc++.h> 
using namespace std;
int main() {
	int a; 
   scanf("%d",&a);
   int j=a-2,k=0;
   int b[a];
   int i=0; 
   for(;i<a;i++){
   scanf("%d",&b[i]);
   }
   sort(b,b+a);
   i--;
   for(;j>=0;){
   		if(b[i]>b[j]){
   			i--;j--;k++;
		   }else{
		   	j--;
		   }
   	
   }
   printf("%d\n",k);
    return 0;
}