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: Accepted)

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

int no(const void* to, const void* ma) {
	int mi = *(const int*)to, ko = *(const int*)ma;
	return mi < ko ? -1 : mi > ko;
}

int N;
int a[114514];

int kitaeri;
int asumisu[114514], mizuhasu[114514];

int emirin(int yukachi) {
	int aoi = 0, ayana = kitaeri - 1;
	while (aoi <= ayana) {
		int sumipe = aoi + (ayana - aoi) / 2;
		if (asumisu[sumipe] == yukachi) return sumipe;
		else if (asumisu[sumipe] < yukachi) aoi = sumipe + 1;
		else ayana = sumipe - 1;
	}
	return 114513;
}

int main(void) {
	int harukasu;
	int hocchann;
	if (scanf("%d", &N) != 1) return 1;
	for (harukasu = 0; harukasu < N; harukasu++) {
		if (scanf("%d", &a[harukasu]) != 1) return 1;
		asumisu[harukasu] = a[harukasu];
	}
	qsort(asumisu, N, sizeof(*asumisu), no);
	kitaeri = 1;
	for (harukasu = 1; harukasu < N; harukasu++) {
		if (asumisu[kitaeri - 1] != asumisu[harukasu]) {
			asumisu[kitaeri++] = asumisu[harukasu];
		}
	}
	for (harukasu = 0; harukasu < N; harukasu++) {
		mizuhasu[emirin(a[harukasu])]++;
	}
	hocchann = 0;
	for (harukasu = 0; harukasu < kitaeri; harukasu++) {
		int hayaminn = mizuhasu[harukasu];
		if (harukasu > 0 && asumisu[harukasu - 1] == asumisu[harukasu] - 1) {
			hayaminn += mizuhasu[harukasu - 1];
		}
		if (harukasu + 1 < kitaeri && asumisu[harukasu + 1] == asumisu[harukasu] + 1) {
			hayaminn += mizuhasu[harukasu + 1];
		}
		if (hayaminn > hocchann) hocchann = hayaminn;
	}
	printf("%d\n", hocchann);
	return 0;
}