Polycarp's Pockets(CodeForces - 1003)

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本文介绍了一个有趣的算法问题：如何用最少的口袋存放不同价值的硬币，确保没有两个相同价值的硬币放在同一个口袋里。通过分析硬币价值的分布情况，我们可以找到最优的解决方案。

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

思路:只需要判断有几个重复的数据就行了,有几个重复的就把他们放到同一个数组里面,,所以最少需要的口袋数就等于出现次数最多的价值。

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <set>

typedef long long LL;
const long long INF = 0x3f3f3f3f;
const long long mod = 1e9+7;
const double PI = acos(-1.0);
const int dir4[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
const int dir8[8][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}, {1, 1}, {-1, -1}, {1, -1}, {-1, 1}};

using namespace std;
int main()
{
    int n,MAX,x;
    int a[110];
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof(a));
        MAX=1;
        for(int i = 0; i < n; i ++)
        {
            scanf("%d",&x);
            a[x] ++;
            MAX = max(MAX, a[x]);
        }
        printf("%d\n",MAX);
    }

    return 0;
}