CodeForces 1011B

Description

Natasha is planning an expedition to Mars for nn people. One of the important tasks is to provide food for each participant.

The warehouse has mm daily food packages. Each package has some food type aiai.

Each participant must eat exactly one food package each day. Due to extreme loads, each participant must eat the same food type throughout the expedition. Different participants may eat different (or the same) types of food.

Formally, for each participant jj Natasha should select his food type bjbj and each day jj-th participant will eat one food package of type bjbj. The values bjbj for different participants may be different.

What is the maximum possible number of days the expedition can last, following the requirements above?

Input

The first line contains two integers nn and mm (1≤n≤1001≤n≤100, 1≤m≤1001≤m≤100) — the number of the expedition participants and the number of the daily food packages available.

The second line contains sequence of integers a1,a2,…,ama1,a2,…,am (1≤ai≤1001≤ai≤100), where aiai is the type of ii-th food package.

Output

Print the single integer — the number of days the expedition can last. If it is not possible to plan the expedition for even one day, print 0.

Sample Input

Input

4 10
1 5 2 1 1 1 2 5 7 2

Output

2

Input

100 1
1

Output

0

Input

2 5
5 4 3 2 1

Output

1

Input

3 9
42 42 42 42 42 42 42 42 42

Output

3

Hint

In the first example, Natasha can assign type 11 food to the first participant, the same type 11 to the second, type 55 to the third and type 22 to the fourth. In this case, the expedition can last for 22 days, since each participant can get two food packages of his food type (there will be used 44 packages of type 11, two packages of type 22 and two packages of type 55).

In the second example, there are 100100 participants and only 11 food package. In this case, the expedition can't last even 11 day.

对存活天数进行暴力。

判断该天数能否成立的条件是，对水果的种类数进行暴力，判断所有的水果一共能支持多少人。

注意STL容器map 的使用。

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #include<stack>
 #include<deque>
 #include<map>
 #include<iostream>
 using namespace std;
 typedef long long  LL;
 const double pi=acos(-1.0);
 const double e=exp();
 const int N = ;
 
 #define lson i << 1,l,m
 #define rson i << 1 | 1,m + 1,r
 
 map<int,int> mp;
 int main()
 {
     int n,m,i,p,j;
     int x;
     scanf("%d%d",&n,&m);
     for(i=;i<=m;i++)
     {
         scanf("%d",&x);
         mp[x]++;
     }
     if(m<n)
         printf("0\n");
     else
     {
         p=m/n;
         map<int,int>::iterator it1;
         for(i=p;i>;i--)
         {
             int sum=;
             for(it1=mp.begin();it1!=mp.end();it1++)
             {
                 sum+=it1->second/i;
             }
             if(sum>=n)
                 break;
         }
         printf("%d\n",i);
     }
     return ;
 }