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>
using namespace std;
#define fi first
#define se second
#define all(a) a.begin(), a.end()
#define pb push_back
#define mp make_pair
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<ll,int> pli;
typedef pair<int,int> pii;
#define rep(i,a,b) for(ll i=a ; i<b ; i++)
#define qrep(que, ite) for(auto ite=begin(que) ; ite!=end(que) ; ite++)
const int max_n = 1e5;
const ll mod = 1e9+7;
const ll INF = 1LL<<60;
const int inf = 1e5;
//typedef long double ld;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
ll mo = 998244353;
struct UnionFind {
    vector<int> par;

    UnionFind(int n) : par(n, -1) { }
    void init(int n) { par.assign(n, -1); }

    int root(int x) {
        if (par[x] < 0) return x;
        else return par[x] = root(par[x]);
    }

    bool issame(int x, int y) {
        return root(x) == root(y);
    }

    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (par[x] > par[y]) swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }

    int size(int x) {
        return -par[root(x)];
    }
};
int main(){
  int n;  cin>>n;
  map<ll,ll> ma;
  rep(i,0,n){
    int x;
    cin >> x;
    ma[x-1]++;
    ma[x]++;
    ma[x+1]++;
  }
  ll ans = 0;
  for(auto j : ma){
    chmax(ans, j.se);
  }
  cout << ans << endl;
  return 0;
}