题目：

A. Mr. Kitayuta, the Treasure Hunter

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The Shuseki Islands are an archipelago of 30001 small islands in the Yutampo Sea. The islands are evenly spaced along a line, numbered from 0 to 30000 from the west to the east. These islands are known to contain many treasures. There are n gems in the Shuseki Islands in total, and the i-th gem is located on island p_i.

Mr. Kitayuta has just arrived at island 0. With his great jumping ability, he will repeatedly perform jumps between islands to the east according to the following process:

• First, he will jump from island 0 to island d.
• After that, he will continue jumping according to the following rule. Let l be the length of the previous jump, that is, if his previous jump was from island prev to island cur, let l = cur - prev. He will perform a jump of length l - 1, l or l + 1 to the east. That is, he will jump to island (cur + l - 1), (cur + l) or (cur + l + 1) (if they exist). The length of a jump must be positive, that is, he cannot perform a jump of length 0 when l = 1. If there is no valid destination, he will stop jumping.

Mr. Kitayuta will collect the gems on the islands visited during the process. Find the maximum number of gems that he can collect.

Input

The first line of the input contains two space-separated integers n and d (1 ≤ n, d ≤ 30000), denoting the number of the gems in the Shuseki Islands and the length of the Mr. Kitayuta's first jump, respectively.

The next n lines describe the location of the gems. The i-th of them (1 ≤ i ≤ n) contains a integer p_i (d ≤ p₁ ≤ p₂ ≤ ... ≤ p_n ≤ 30000), denoting the number of the island that contains the i-th gem.

Output

Print the maximum number of gems that Mr. Kitayuta can collect.

Examples

Input

Copy

4 10
10
21
27
27

Output

Copy

3

Input

Copy

8 8
9
19
28
36
45
55
66
78

Output

Copy

6

Input

Copy

13 7
8
8
9
16
17
17
18
21
23
24
24
26
30

Output

Copy

4

Note

In the first sample, the optimal route is 0  →  10 (+1 gem)  →  19  →  27 (+2 gems)  → ...

In the second sample, the optimal route is 0  →  8  →  15  →  21 →  28 (+1 gem)  →  36 (+1 gem)  →  45 (+1 gem)  →  55 (+1 gem)  →  66 (+1 gem)  →  78 (+1 gem)  → ...

In the third sample, the optimal route is 0  →  7  →  13  →  18 (+1 gem)  →  24 (+2 gems)  →  30 (+1 gem)  → ...

思路和实现都不难的动态规划

(._. )

做的时候没看出来长度的限制 担心复杂度太大所以不敢写

(._. )

菜鸡

(._. )

dp[i][j]表示到编号为 i 的岛且上次跳跃长度为 j 时能取到的最多的gems的数目。

注意：岛的编号限制在30000以内，且每次最多增长一步，第一步跳跃长度为d，总的跳跃长度 = d + (d + 1) + (d + 2) + ... + (d + 245)  ≥  1 + 2 + ... + 245 = 245·(245 + 1) / 2  =  30135  >  30000。所以跳跃长度最长为（d+245），最短为（d-245），因此 j 的枚举长度在[d-245，d+245]之间，第二维的空间缩小到500。

#include<iostream>

#include<cstring>

#include<cstdio>

#include<algorithm>

#include<set>

using namespace std;

const int maxn = *1e4+;

int dp[maxn][];;

int num[maxn];

int main()

{

    int base;

    int a, b;

    int n, m, d;

    int i, j, k;

    scanf("%d %d",&n,&d);

    for(i = ; i <= n; i++)

        scanf("%d", &a), num[a]++;

    base = max(d -,);

    memset(dp, -, sizeof dp);

    dp[d][d-base] = num[d];

    int ans=;

    for(i = d ; i <= ; i++)

    {

        for( j =  ;j <= ; j++)

        {

            if(dp[i][j] == -) continue;

            for(k = -; k < ; k++)

            {

                if( j + k < || base + i + j + k > ) continue;

                dp[i + base + j + k][j + k] = max(dp[base + i + j + k][j + k], dp[i][j] + num[i + j + k + base]);

            }

            ans = max(ans, dp[i][j]);

        }

    }

    cout << ans << endl;

    return ;

}