Problem: There is a set consisting of N distinct integers. The i-th smallest element in this set is S_i. We want to divide this set into two sets, X and Y, such that:

* The absolute difference of any two distinct elements in X is A or greater.
* The absolute difference of any two distinct elements in Y is B or greater.



How many ways are there to perform such division, modulo 10^9 + 7? Note that one of X and Y may be empty.

Constraints

* All input values are integers.
* 1 ≦ N ≦ 10^5
* 1 ≦ A , B ≦ 10^{18}
* 0 ≦ S_i ≦ 10^{18}(1 ≦ i ≦ N)
* S_i < S_{i+1}(1 ≦ i ≦ N - 1)

Input

The input is given from Standard Input in the following format:


N A B
S_1
:
S_N


Output

Print the number of the different divisions under the conditions, modulo 10^9 + 7.

Examples

Input

5 3 7
1
3
6
9
12


Output

5


Input

7 5 3
0
2
4
7
8
11
15


Output

4


Input

8 2 9
3
4
5
13
15
22
26
32


Output

13


Input

3 3 4
5
6
7


Output

0

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ls(x) x<<1
#define rs(x) x<<1|1
const int MN=100000+5;
const int P=1000000007;
using namespace std;
typedef long long ll;
template<typename T>inline T IN(T&in){
	in=0;char c=getchar();int f=1;
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c))in=in*10+c-'0',c=getchar();
	return in*=f;
}
int n;ll A,B;
ll f[MN],a[MN],t[MN*4];
void mod(int rt,int l,int r,int p,int d){
	t[rt]=(t[rt]+d)%P;
	if(l==r)return ;
	int mid=l+r>>1;
	if(p<=mid)mod(ls(rt),l,mid,p,d);
	else mod(rs(rt),mid+1,r,p,d);
}
int ask(int rt,int l,int r,int L,int R){
	if(L>R)return 0;
	if(L<=l&&r<=R)return t[rt];
	int mid=l+r>>1,res=0;
	if(L<=mid)res=(res+ask(ls(rt),l,mid,L,R))%P;
	if(mid+1<=R)res=(res+ask(rs(rt),mid+1,r,L,R))%P;
	return res;
}
void input(){
	IN(n),IN(A),IN(B);if(A<B)swap(A,B);
	for(int i=1;i<=n;++i)IN(a[i]);
	a[0]=-A,a[n+1]=a[n]+A;mod(1,0,n,0,f[0]=1);
	int l=0,r=0;
	for(int i=1;i<=n+1;i++){
		while(a[i]-a[r+1]>=A)r++;
		if(i>=2&&a[i-1]-a[i-2]<B)l=i-2;
		if(a[i]-a[i-1]>=A)f[i]=f[i-1];
		f[i]=(f[i]+ask(1,0,n,l,min(r,i-2)))%P;
		if(i<=n&&a[i+1]-a[i-1]>=B)mod(1,0,n,i,f[i]);
	}
	printf("%lld\n",f[n+1]);
}
int main(){
	
	input();
	return 0;
}