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 <bits/stdc++.h>
#define For(i, a, b) for(int (i)=(int)(a); (i)<(int)(b); ++(i))
#define rFor(i, a, b) for(int (i)=(int)(a)-1; (i)>=(int)(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
#define fi first
#define se second
using namespace std;
typedef long long lint;
typedef unsigned long long ulint;
typedef pair<int, int> pii;
typedef pair<lint, lint> pll;
template<class T> bool chmax(T &a, const T &b){if(a<b){a=b; return true;} return false;}
template<class T> bool chmin(T &a, const T &b){if(a>b){a=b; return true;} return false;}
template<class T> T div_floor(T a, T b){
    if(b < 0) a *= -1, b *= -1;
    return a>=0 ? a/b : (a+1)/b-1;
}
template<class T> T div_ceil(T a, T b){
    if(b < 0) a *= -1, b *= -1;
    return a>0 ? (a-1)/b+1 : a/b;
}

constexpr lint mod = 1000000007;
constexpr lint INF = mod * mod;
constexpr int MAX = 200010;

template<int_fast64_t MOD> struct modint{
    using i64=int_fast64_t;
    i64 a;
    modint(const i64 a_=0): a(a_){
        if(a>MOD) a%=MOD;
        else if(a<0) (a%=MOD)+=MOD;
    }
    modint inv(){
        i64 t=1, n=MOD-2, x=a;
        while(n){
            if(n&1) (t*=x)%=MOD;
            (x*=x)%=MOD;
            n>>=1;
        }
        modint ret(t);
        return ret;
    }
    bool operator==(const modint x) const{return a==x.a;}
    bool operator!=(const modint x) const{return a!=x.a;}
    modint operator+(const modint x) const{
        return modint(*this)+=x;
    }
    modint operator-(const modint x) const{
        return modint(*this)-=x;
    }
    modint operator*(const modint x) const{
        return modint(*this)*=x;
    }
    modint operator/(const modint x) const{
        return modint(*this)/=x;
    }
    modint operator^(const lint x) const{
        return modint(*this)^=x;
    }
    modint &operator+=(const modint &x){
        a+=x.a;
        if(a>=MOD) a-=MOD;
        return *this;
    }
    modint &operator-=(const modint &x){
        a-=x.a;
        if(a<0) a+=MOD;
        return *this;
    }
    modint &operator*=(const modint &x){
        (a*=x.a)%=MOD;
        return *this;
    }
    modint &operator/=(modint x){
        (a*=x.inv().a)%=MOD;
        return *this;
    }
    modint &operator^=(lint n){
        i64 ret=1;
        while(n){
            if(n&1) (ret*=a)%=MOD;
            (a*=a)%=MOD;
            n>>=1;
        }
        a=ret;
        return *this;
    }
    modint operator-() const{
        return modint(0)-*this;
    }
    modint &operator++(){
        return *this+=1;
    }
    modint &operator--(){
        return *this-=1;
    }
    bool operator<(const modint x) const{
        return a<x.a;
    }
};

using mint=modint<1000000007>;

vector<mint> fact;
vector<mint> revfact;

void setfact(int n){
    fact.resize(n+1); revfact.resize(n+1);
    fact[0] = 1;
    rep(i, n) fact[i+1] = fact[i] * mint(i+1);

    revfact[n] = fact[n].inv();
    for(int i=n-1; i>=0; i--) revfact[i] = revfact[i+1] * mint(i+1);
}

mint getC(int n, int r){
    if(n<r) return 0;
    return fact[n] * revfact[r] * revfact[n-r];
}

int main(){
    int n, t;
    scanf("%d%d", &n, &t);
    int a[n];
    rep(i, n) scanf("%d", &a[i]);
    sort(a, a+n);
    mint ans = 1;
    rep(i, n){
        ans *= (mint)(upper_bound(a, a+n, a[i] + t) - (a+i));
    }
    printf("%lld\n", ans.a);
}