Problem: 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 pi.

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 pi (d ≤ p1 ≤ p2 ≤ ... ≤ pn ≤ 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

4 10
10
21
27
27


Output

3


Input

8 8
9
19
28
36
45
55
66
78


Output

6


Input

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


Output

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)  → ...

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
T max2(T a, T b) {
  return a < b ? b : a;
}
template <class T>
T min2(T a, T b) {
  return a < b ? a : b;
}
template <class T>
T max3(T a, T b, T c) {
  return max2(max2(a, b), c);
}
template <class T>
T min3(T a, T b, T c) {
  return min2(min2(a, b), c);
}
template <class T>
T max4(T a, T b, T c, T d) {
  return max2(max2(a, b), max2(c, d));
}
template <class T>
T min4(T a, T b, T c, T d) {
  return min2(min2(a, b), max2(c, d));
}
const long long int mod = 1000000007ll;
const long long int MOD = 1000000009ll;
long long int bit_count(long long int _x) {
  long long int _ret = 0;
  while (_x) {
    if (_x % 2 == 1) _ret++;
    _x /= 2;
  }
  return _ret;
}
long long int bit(long long int _mask, long long int _i) {
  long long int x = 1;
  return (_mask & (x << _i)) == 0 ? 0 : 1;
}
long long int powermod(long long int _a, long long int _b,
                       long long int _m = mod) {
  long long int _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a) % _m;
    _b /= 2;
    _a = (_a * _a) % _m;
  }
  return _r;
}
long long int power(long long int _a, long long int _b) {
  long long int _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a);
    _b /= 2;
    _a = (_a * _a);
  }
  return _r;
}
long long int add(long long int a, long long int b, long long int m = mod) {
  long long int x = a + b;
  while (x >= m) x -= m;
  return x;
}
long long int sub(long long int a, long long int b, long long int m = mod) {
  long long int x = a - b;
  while (x < 0) x += m;
  return x;
}
long long int mul(long long int a, long long int b, long long int m = mod) {
  long long int x = a * b;
  x %= m;
  return x;
}
long long int gcd(long long int a, long long int b) {
  if (a < b) {
    a = a ^ b;
    b = a ^ b;
    a = a ^ b;
  }
  if (b == 0) return a;
  long long int t;
  while (b) {
    t = b;
    b = a % b;
    a = t;
  }
  return a;
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
vector<string> _split(const string& second, char c) {
  vector<string> v;
  stringstream ss(second);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return v;
}
pair<int, int> operator+(const pair<int, int>& l, const pair<int, int>& r) {
  pair<int, int> ret;
  ret.first = l.first + r.first;
  ret.second = l.second + r.second;
  return ret;
}
struct cmp {
  bool operator()(long long int const& l, long long int const& r) {
    return l > r;
  }
} myobject;
int dp[30010][510];
int first[30010];
int n, d, p;
int solve(int i, int j, int l) {
  if (i > 30000) return 0;
  if (dp[i][j] != -1) return dp[i][j];
  dp[i][j] = 0;
  if (l >= 2)
    dp[i][j] =
        first[i] + max3<int>(solve(i + l, j, l), solve(i + l + 1, j + 1, l + 1),
                             solve(i + l - 1, j - 1, l - 1));
  else if (l >= 1)
    dp[i][j] =
        first[i] + max(solve(i + l, j, l), solve(i + l + 1, j + 1, l + 1));
  return dp[i][j];
}
int main() {
  {
    scanf("%d", &n);
    ;
    scanf("%d", &d);
    ;
  };
  fill(first, first + 30010, 0);
  for (int i = 0; i < 30010; i++) fill(dp[i], dp[i] + 510, -1);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p);
    ;
    first[p]++;
  }
  int ans = first[0];
  ans += solve(d, 250, d);
  cout << ans;
  return 0;
}