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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include<algorithm>
using namespace std;

template <class T> using V = vector<T>;

using ll = long long;
using db = double;
using st = string;
using ch = char;
using vll = V<ll>;
using vpll =V<pair<ll,ll>>;
using vst = V<st>;
using vdb = V<db>;
using vch = V<ch>;


#define FOR(i,a,b) for(ll i=(a);i<(ll)(b);i++)
#define rFOR(i,a,b) for(ll i=(b);i>(ll)(a);i--)
#define oFOR(i,a,b) for(ll i=(a);i<(ll)(b);i+=2)
#define bgn begin()
#define en end()
#define SORT(a) sort((a).bgn,(a).en)
#define REV(a) reverse((a).bgn,(a).en)
#define M(a,b) max(a,b)
#define rM(a,b) min(a,b)
#define fi first
#define se second
#define sz size()
#define gcd(a,b) __gcd(a,b)
#define co(a) cout<<a<<endl;
#define ci(a) cin>>a;


ll sum(ll n) {
  ll m=0;
  FOR(i,0,20){
    m+=n%10;
    n/=10;
    if(n==0){
      break;
    }
  }
  return m;
}


ll combi(ll n,ll m) {
  ll ans=1;
  rFOR(i,n,n-m){
    ans*=i;
  }
  FOR(i,1,m+1){
    ans/=i;
  }
  return ans;
}


ll lcm(ll a,ll b){
  ll n;
  n=a/gcd(a,b)*b;
  return n;
}


/****************************************\
| Thank you for viewing my code:)        |
| Written by RedSpica a.k.a. RanseMirage |
| Twitter:@asakaakasaka                  | 
\****************************************/


signed main() {
  ll n;
  	ci(n);
  vll A(3*n);
  ll m=0;
  for(int i=0;i<3*n;i+=3){
    ci(A[i])
    A[i+1]=A[i]-1;
    A[i+2]=A[i]+1;
  }
  SORT(A);
  ll count=1;
  ll ans=1;
  FOR(i,1,3*n){
    if(A[i-1]==A[i]){
      count++;
    }
    else{
      ans=M(count,ans);
      count=1;
    }
  }
  ans=M(count,ans);
  co(ans)
}