Problem: Honestly, a rabbit does not matter.

There is a rabbit playing a stage system action game. In this game, every stage has a difficulty level. The rabbit, which always needs challenges, basically wants to play more difficult stages than he has ever played. However, he sometimes needs rest too. So, to compromise, he admitted to play T or less levels easier stages than the preceding one.

How many ways are there to play all the stages at once, while honoring the convention above? Maybe the answer will be a large number. So, let me know the answer modulo 1,000,000,007.



Input

The first line of input contains two integers N and T (1 ≤ N ≤ 100,000, 1 ≤ T ≤ 10,000). N is the number of stages, and T is the compromise level.

The following N lines describe the difficulty levels of each stage. The i-th line contains one integer Di (1 ≤ Di ≤ 100,000), which is the difficulty level of the i-th stage.

Output

Calculate how many ways to play all the stages at once there are. Print the answer modulo 1,000,000,007 in a line.

Examples

Input

3 1
1
2
3


Output

4


Input

5 3
9
2
6
8
8


Output

24


Input

5 7
9
9
9
1
5


Output

48

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <cfloat>
#include <ctime>
#include <cassert>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <numeric>
#include <list>
#include <iomanip>
#include <fstream>
#include <bitset>
   
using namespace std;
   
 
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define foreach(it, c) for (__typeof__((c).begin()) it=(c).begin(); it != (c).end(); ++it)
#define rforeach(it, c) for (__typeof__((c).rbegin()) it=(c).rbegin(); it != (c).rend(); ++it)
#define all(c) (c).begin(), (c).end()
#define rall(c) (c).rbegin(), (c).rend()
#define CL(arr, val) memset(arr, val, sizeof(arr))
#define COPY(dest, src) memcpy(dest, src, sizeof(dest))
#define ten(n) ((long long)(1e##n))
#define bin(n) (1LL << (n))
#define erep(i, n) for (int i = 0; i <= (int)(n); ++i)
#define revrep(i, n) for (int i = (n); i >= 0; --i)
#define pb push_back

template <class T> void chmax(T& a, const T& b) { a = max(a, b); }
template <class T> void chmin(T& a, const T& b) { a = min(a, b); }

template <class T> void uniq(T& c) { sort(c.begin(), c.end()); c.erase(unique(c.begin(), c.end()), c.end()); }
 
template <class T> string to_s(const T& a) { ostringstream os; os << a; return os.str(); }
template <class T> T to_T(const string& s) { istringstream is(s); T res; is >> res; return res; }

template <typename T> void print_container(ostream& os, const T& c) { const char* _s = " "; if (!c.empty()) { __typeof__(c.begin()) last = --c.end(); foreach (it, c) { os << *it; if (it != last) cout << _s; } } }
template <typename T> ostream& operator<<(ostream& os, const vector<T>& c) { print_container(os, c); return os; }
template <typename T> ostream& operator<<(ostream& os, const set<T>& c) { print_container(os, c); return os; }
template <typename T> ostream& operator<<(ostream& os, const multiset<T>& c) { print_container(os, c); return os; }
template <typename T> ostream& operator<<(ostream& os, const deque<T>& c) { print_container(os, c); return os; }
template <typename T, typename U> ostream& operator<<(ostream& os, const map<T, U>& c) { print_container(os, c); return os; }
template <typename T, typename U> ostream& operator<<(ostream& os, const pair<T, U>& p) { os << "( " << p.first << ", " << p.second << " )"; return os; }

template <class T> void print(T a, int n, const string& deli = " ", int br = 1) { for (int i = 0; i < n; ++i) { cout << a[i]; if (i + 1 != n) cout << deli; } while (br--) cout << endl; }
template <class T> void print2d(T a, int w, int h, int width = -1, int br = 1) { for (int i = 0; i < h; ++i) { for (int j = 0; j < w; ++j) {    if (width != -1) cout.width(width); cout << a[i][j] << ' '; } cout << endl; } while (br--) cout << endl; }

template <class T> void input(T& a, int n) { for (int i = 0; i < n; ++i) cin >> a[i]; }
template <class T> void input(T* a, int n) { for (int i = 0; i < n; ++i) cin >> a[i]; }

void fix_pre(int n) { cout.setf(ios::fixed, ios::floatfield); cout.precision(10); }
void fast_io() { cin.tie(0); ios::sync_with_stdio(false); }
#define trace(x) (cout << #x << ": " << (x) << endl)
 
bool in_rect(int x, int y, int w, int h) { return 0 <= x && x < w && 0 <= y && y < h; }


typedef long long ll;
typedef pair<int, int> pint;
 
// y(v): v>^<  y(^): ^>v<
const int dx[] = { 0, 1, 0, -1 };
const int dy[] = { 1, 0, -1, 0 };

const double PI = acos(-1.0);



int main()
{
    int n, t, d[ten(5)];
    cin >> n >> t;
    input(d, n);
    sort(d, d + n);

    const ll mod = ten(9) + 7;
    ll res = 1;
    int lower = 0;
    rep(i, n)
    {
        while (d[lower] < d[i] - t)
            ++lower;
        (res *= i - lower + 1) %= mod;
    }
    cout << res << endl;
}