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], we can permute it so that it becomes [20,40,10,30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. 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 n (1≤n≤105) — the length of the array. The second line contains n integers a 1,a2…,an (1≤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]. 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.

思路

對於陣列中的每一個位置，將比該位置原來的數大並且儘可能小的數放在該位置 先用sort排一下序，對於每個數，找到後面比他大的第一個數就行。注意不要多放。

#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <cstdio>
#include <vector>
#include <utility>
#include <cstring>
#include <iostream>
#include <algorithm>
#define eps 1e-8
#define PI acos(-1)
#define INF 0x3f3f3f3f
#define N 100005
#define newmax(a,b) a>b?a:b
#define newmin(a,b) a>b?b:a
#define Lowbit(x) (x&-x)
using namespace std;
typedef long long int LL;
const int dir[4][2]= { {1,0},{0,1},{-1,0},{0,-1} };


int main()
{
    int n,ans=0;
    int a[N],b[N];
    cin>>n;
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
    sort(a,a+n);
    for(int i=0;i<n;i++)
        b[i]=a[i];
    int p=0;
    for(int i=0;i<n;)
    {
        if(p>=n)
            break;
        if(b[p]>a[i])
        {
            ans++;
            i++;
            p++;
        }
        else
            p++;
    }
    printf("%d\n",ans);
    return 0;
}