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>
using namespace std;
#define int long long
const int N=100005,M=1e9+7;
struct Tree{
	int l,r,num;
}T[N*4];
int n,x,y,a[N],f[N],g[N],b[N];
void build(int x,int l,int r){
	T[x].l=l;T[x].r=r;
	if (l==r){
		T[x].num=b[l];
		return;
	}
	int mid=(l+r)/2;
	build(x*2,l,mid);
	build(x*2+1,mid+1,r);
	T[x].num=min(T[x*2].num,T[x*2+1].num);
}
int find(int x,int l,int r){
	if (l>r)return 1e18;
	if (T[x].l>r||l>T[x].r)return 1e18;
	if (T[x].l>=l&&T[x].r<=r)return T[x].num;
	return min(find(x*2,l,r),find(x*2+1,l,r));
}
signed main(){
	scanf("%lld%lld%lld",&n,&x,&y);
	for (int i=1;i<=n;i++)scanf("%lld",&a[i]);
	a[0]=-1e18;
	for (int i=1;i<n;i++)b[i]=a[i]-a[i-1];
	build(1,1,n);
	f[1]=g[1]=1;
	for (int i=2;i<=n;i++){
		int l=0,r=i-1;
		while (l<r){
			int mid=(l+r+1)/2;
			if (a[i]-a[mid]>=x)l=mid;
			else r=mid-1; 
		}
		if (i==l+1)f[i]=(f[i-1]+g[i-1])%M; 
		else if (find(1,l+2,i-1)>=y)f[i]=g[l+1];
		l=0,r=i-1;
		while (l<r){
			int mid=(l+r+1)/2;
			if (a[i]-a[mid]>=y)l=mid;
			else r=mid-1; 
		}
		if (i==l+1)g[i]=(g[i-1]+f[i-1])%M; 
		else if (find(1,l+2,i-1)>=x)g[i]=f[l+1];
	} 
	printf("%lld\n",(f[n]+g[n])%M);
}