Polycarp's Pockets (cf 1003A)

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解决分配不同价值的硬币到两个口袋中，确保每个口袋内没有相同价值的硬币，以找到最少需要的口袋数。通过输入硬币的数量和价值，输出所需的最少口袋数。

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Polycarp has nn coins, the value of the ii-th coin is aiai. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.

For example, if Polycarp has got six coins represented as an array a=[1,2,4,3,3,2]a=[1,2,4,3,3,2], he can distribute the coins into two pockets as follows: [1,2,3],[2,3,4][1,2,3],[2,3,4].

Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that.

Input

The first line of the input contains one integer nn (1≤n≤1001≤n≤100) — the number of coins.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100) — values of coins.

Output

Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket.

Examples

input

6
1 2 4 3 3 2

output

input

1
100

output

代码如下：

#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <ctime>
#define maxn 10007
#define N 107
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define eps 0.000000001
#define mod 1000000007
using namespace std;
typedef long long ll;

int main()
{
    int n;
    cin>>n;
    int a[111];
    for(int i=0; i<n; i++)
    {
        int m;
        cin>>m;
        a[m]++;
    }
    int max1=-INF;
    for(int i=1; i<=100; i++)
        max1=max(max1,a[i]);
    cout<<max1<<endl;
    return 0;
}