Problem:
Score : 300 points

Problem Statement

You are given an integer sequence of length N, a_1,a_2,...,a_N.

For each 1≤i≤N, you have three choices: add 1 to a_i, subtract 1 from a_i or do nothing.

After these operations, you select an integer X and count the number of i such that a_i=X.

Maximize this count by making optimal choices.

Constraints

1≤N≤10^5

0≤a_i<10^5 (1≤i≤N)

a_i is an integer.

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 .. a_N

Output

Print the maximum possible number of i such that a_i=X.

Sample Input 1

7
3 1 4 1 5 9 2

Sample Output 1

4

For example, turn the sequence into 2,2,3,2,6,9,2 and select X=2 to obtain 4, the maximum possible count.

Sample Input 2

10
0 1 2 3 4 5 6 7 8 9

Sample Output 2

3

Sample Input 3

1
99999

Sample Output 3

1

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_poizcy.hpp>
typedef long long  ll;
#define pb                push_back
#define mp                make_pair
#define all(a)            (a).begin(), (a).end()
#define mem(a,h)          memset(a, (h), sizeof(a))
#define fi first
#define se second
#define por(a,b) (((a%MOD) * (b%MOD))%MOD)
#define forg(i, b, e, c)    for (ll i = (ll)b; i < (ll)e; i+=c)
#define forr(i, b, e)    for (int i = b; i < e; i++)
 
 
using namespace std;
//using namespace __gnu_pbds;
typedef double lldb;
typedef pair<ll, ll>  ii;
typedef pair<double, double>  iidb;
#define iii pair<ii, nodo*>
typedef vector<ll>     vi;
typedef vector<vi>     vvi;
typedef vector<ii>      vii;
typedef vector<ll>      vll;
//typedef tree<ii,null_type,less<ii>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
const ll INF = 1e9+7;
const double PI = acos(-1);
#define initseg ll new_nodo=(pnodo*2),mid=(iz+der)/2;
 
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define tam 3000
#define offset 3000
#define ptr nodo*
const int MOD=1e9+7;

using namespace std;

int main()
{	
	int n;
	cin>>n;
	map<int,int> m;
	forr(i,0,n)
	{
		int valor;
		cin>>valor;
		m[valor-1]++;
		m[valor+1]++;
		m[valor]++;
	}
	ii maxx={0,0};
	for(auto par:m)
	{
		maxx=max(maxx,{par.se,par.fi});
	}
	cout<<maxx.fi<<endl;
}