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<stdio.h>
#include<string.h>
#include<algorithm>
#define N 100010
#define LL long long
#define M 1000000007
using namespace std;
int n,m,f[N]; LL A,B,s[N],w[N<<2];
void change(int p,int l,int r,int a,int b){
    if (l==r){
        (w[p]+=b)%=M;
        return;
    }
    int mid=(l+r)/2;
    if (a<=mid) change(p*2,l,mid,a,b);else change(p*2+1,mid+1,r,a,b);
    w[p]=(w[p*2]+w[p*2+1])%M;
}
int query(int p,int l,int r,int a,int b){
    if (a>b) return 0;
    if (l==a&&r==b) return w[p];
    int mid=(l+r)/2;
    if (b<=mid) return query(p*2,l,mid,a,b);
    else if (a>mid) return query(p*2+1,mid+1,r,a,b);
    else return (query(p*2,l,mid,a,mid)+query(p*2+1,mid+1,r,mid+1,b))%M;
}
int main(){
	scanf("%d%lld%lld",&n,&A,&B);
	if(A<B) swap(A,B); f[0]=1; 
	for(int i=1;i<=n;++i) scanf("%lld",s+i);
	*s=-A; s[n+1]=A+s[n]; int l=0,r=0; change(1,0,n,0,f[0]);
	for(int i=1;i<=n+1;++i){
		while(s[i]-s[r+1]>=A) ++r;
		if(i>=2 && s[i-1]-s[i-2]<B) l=i-2;
		if(s[i]-s[i-1]>=A) f[i]=f[i-1];
		f[i]=(f[i]+query(1,0,n,l,min(r,i-2)))%M;
		if(s[i+1]-s[i-1]>=B) change(1,0,n,i,f[i]);
	}
	printf("%lld\n",(f[n+1]+M)%M);
}