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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}

int main(void){
	int n;
	scanf("%d",&n);
	int a[n];
	for(int i=0;i<n;i++) {scanf("%d",&a[i]);}
	qsort(a, n, sizeof(int), compare_int);//昇順並び替え
	//for(int i=0;i<n;i++) {printf("%d",a[i]);}
	
	int before=0;
	int now=0;
	int beforebefore=0;
	int temp=-1;
	int countmax=0;
	for(int i=0;i<n;i++){
		if(temp==a[i]){
			now++;
		}
		else{
			temp=a[i];
			if(countmax<beforebefore+before+now){countmax=beforebefore+before+now;}
			//printf("%d:%d,%d,%d\n",a[i],beforebefore,before,now);
			beforebefore=before;before=now;now=1;
		}
	}
	if(countmax<beforebefore+before+now){countmax=beforebefore+before+now;}
	//printf("last:%d,%d,%d\n",beforebefore,before,now);
	printf("%d",countmax);
	return 0;
}