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: Time Limit Exceeded)

#include<stdio.h>
#include<math.h>
void px(int a[],int n){
	for(int j=1;j<n;j++){
		int l=j;
		for(int k=j;k<=n;k++){
			if(a[l]>a[k])l=k;
		}
		int t=a[l];
		a[l]=a[j];
		a[j]=t;
	}
}
int main(){
	int n,a[100000+5],b[100000+5];
	while(~scanf("%d",&n)){
		if(n==1){
			printf("1\n");
			continue;
		}
		int num=0;
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
		}
		px(a,n);
		int numa=2,numb=1,begin=0;
		for(;;){
			int num=0;
			for(int i=numa;i<=n;i++){
				if(fabs(a[i]-a[i-1])==1){
					numa=i;
					begin++;
					break;
				}
				begin++;
			}
			for(int i=1;i<=n;i++){
				if(a[i]+1==a[numa] || a[i]-1==a[numa] || a[i]==a[numa]){
					num++;
				}
			}
			b[numb]=num;
			if(begin==n)break;
			numb++;
		}
		px(b,numb);
		printf("%d\n",b[numb]);
	}
}