903CBoxes Packing

C. Boxes Packing

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mishka has got n empty boxes. For every i (1 ≤ i ≤ n), i-th box is a cube with side length ai.

Mishka can put a box i into another box j if the following conditions are met:

• i-th box is not put into another box;
• j-th box doesn't contain any other boxes;
• box i is smaller than box j (ai < aj).

Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff it is not put into some another box.

Help Mishka to determine the minimum possible number of visible boxes!

Input

The first line contains one integer n (1 ≤ n ≤ 5000) — the number of boxes Mishka has got.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the side length of i-th box.

Output

Print the minimum possible number of visible boxes.

Examples

input

31 2 3

output

1

input

44 2 4 3

output

2

Note

In the first example it is possible to put box 1 into box 2, and 2 into 3.

In the second example Mishka can put box 2 into box 3, and box 4 into box 1.

题意：给你n个盒子，把小盒子放进大的盒子里，这样只能看到大盒子，输出最小能够有看到的盒子数。


题解：最开始想的是排序，前面一个比后面小就能放，否则不能放，输出盒子个数加1，但是这样不仅时间会超，而且有思维漏洞，
10
86 89 89 86 86 89 86 86 89 89这组数据那样算，输出9，而正确答案是5，因为相同的盒子可以放进比它大的盒子里，再放进大盒子里。所以用map处理标记每个盒子
出现次数，取最大的也是结果。

#include<bits/stdc++.h>using namespace std;map<int,int>m;int main(){    int n,ans,a;    while(cin>>n)    {        ans=0;        m.clear();        for(int i=0; i<n; i++)        {            cin>>a;            m[a]++;            ans=max(m[a],ans);        }        cout<<ans<<endl;    }    return 0;}