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>
                            #include<iostream>
                            #include<cstdio>
                            #include<vector>
                            #include<queue>
                            #include<map>
                            #include<cstring>
                            #include<string>
                            #include <math.h>
                            #include<algorithm>
                        //    #include <boost/multiprecision/cpp_int.hpp>
                            #include<functional>
                    #define int long long
                            #define inf  1000000007
                            #define pa pair<int,int>
                            #define ll long long
                            #define pal pair<double,pa>
                            #define ppa pair<int,int>
                            #define ppap pair<int,pa>
                            #define ssa pair<string,int>
                            #define  mp make_pair
                            #define  pb push_back
                            #define EPS (1e-10)
                            #define equals(a,b) (fabs((a)-(b))<EPS)
                     
                            using namespace std;
                  
                            class Point{
                            	public:
                            	double x,y;
                            	Point(double x=0,double y=0):x(x),y(y) {}
                            	Point operator + (Point p) {return Point(x+p.x,y+p.y);}
                            	Point operator - (Point p) {return Point(x-p.x,y-p.y);}
                            	Point operator * (double a) {return Point(x*a,y*a);}
                            	Point operator / (double a) {return Point(x/a,y/a);}
                            	double absv() {return sqrt(norm());}
                            	double norm() {return x*x+y*y;}
                            	bool operator < (const Point &p) const{
                            		return x != p.x ? x<p.x: y<p.y;
                            	}
                            	bool operator == (const Point &p) const{
                            		return fabs(x-p.x)<EPS && fabs(y-p.y)<EPS;
                            	}
                            };
                            typedef Point Vector;
                     
                            struct Segment{
                            Point p1,p2;
                            };
                             double dot(Vector a,Vector b){
                            	return a.x*b.x+a.y*b.y;
                            }
                            double cross(Vector a,Vector b){
                            	return a.x*b.y-a.y*b.x;
                            }
                        
bool parareru(Point a,Point b,Point c,Point d){
//	if(abs(cross(a-b,d-c))<EPS)cout<<"dd "<<cross(a-b,d-c)<<endl;
	return abs(cross(a-b,d-c))<EPS;
}
double distance_ls_p(Point a, Point b, Point c) {
  if ( dot(b-a, c-a) < EPS ) return (c-a).absv();
  if ( dot(a-b, c-b) < EPS ) return (c-b).absv();
  return abs(cross(b-a, c-a)) / (b-a).absv();
}
bool is_intersected_ls(Segment a,Segment b) {
	if(a.p1==b.p1||a.p2==b.p1||a.p1==b.p2||a.p2==b.p2) return false;
	if(parareru((a.p2),(a.p1),(a.p1),(b.p2))&&parareru((a.p2),(a.p1),(a.p1),(b.p1))){
//		cout<<"sss"<<endl;
		if(dot(a.p1-b.p1,a.p1-b.p2)<EPS) return true;
		if(dot(a.p2-b.p1,a.p2-b.p2)<EPS) return true;
		if(dot(a.p1-b.p1,a.p2-b.p1)<EPS) return true;
		if(dot(a.p1-b.p2,a.p2-b.p2)<EPS) return true;
		return false;
	}
  else return ( cross(a.p2-a.p1, b.p1-a.p1) * cross(a.p2-a.p1, b.p2-a.p1) < EPS ) && ( cross(b.p2-b.p1, a.p1-b.p1) * cross(b.p2-b.p1, a.p2-b.p1) < EPS );
}

double segment_dis(Segment a,Segment b){
	if(is_intersected_ls(a,b))return 0;
	double r=distance_ls_p(a.p1, a.p2, b.p1);
	r=min(r,distance_ls_p(a.p1, a.p2, b.p2));
	r=min(r,distance_ls_p(b.p1, b.p2, a.p2));
	r=min(r,distance_ls_p(b.p1, b.p2, a.p1));
	return r;
}
Point intersection_ls(Segment a, Segment b) {
  Point ba = b.p2-b.p1;
  double d1 = abs(cross(ba, a.p1-b.p1));
  double d2 = abs(cross(ba, a.p2-b.p1));
  double t = d1 / (d1 + d2);

  return a.p1 + (a.p2-a.p1) * t;
}

                string itos( int i ) {
                ostringstream s ;
                s << i ;
                return s.str() ;
                }
                 
                int gcd(int v,int b){
                	if(v>b) return gcd(b,v);
                	if(v==b) return b;
                	if(b%v==0) return v;
                	return gcd(v,b%v);
                }
 
                double distans(double x1,double y1,double x2,double y2){
                	double rr=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
                	return sqrt(rr);
                	
                }
                /*
int pr[100010];
//int inv[100010];
*/ 
int beki(int wa,int rr,int warukazu){
	if(rr==0) return 1ll;
	if(rr==1) return wa%warukazu;
	if(rr%2==1) return (beki(wa,rr-1,warukazu)*wa)%warukazu;
	int zx=beki(wa,rr/2,warukazu);
	return (zx*zx)%warukazu;
}
  /*
void gya(){
	pr[0]=1;
	for(int i=1;i<100010;i++){
		pr[i]=(pr[i-1]*i)%inf;
	}
	for(int i=0;i<100010;i++) inv[i]=beki(pr[i],inf-2);
	
}

*/

                         //----------------kokomade tenpure------------
int sum[100020]={0};
int a[100002];
signed main(){
	
int n,t;
	cin>>n>>t;
	for(int i=0;i<n;i++) {
		cin>>a[i];
		sum[a[i]]++;
	}
	for(int i=1;i<=100010;i++) sum[i] += sum[i-1];
	
	int ans=1;
	sort(a,a+n);
	for(int i=0;i<n;i++){
		int r=0;
		if(a[i]-t>1) r=sum[a[i]-t-1];
		r=i-r+1;
	//	cout<<r<<endl;
		ans*=r;
		ans %=inf;
	}
	cout<<ans<<endl;
	
	
	
	
	
	
	return 0;
}