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<cstdio>
#include<algorithm>
#define maxn 100005
#define LL long long
#define mod 1000000007
#define INF (LL)4e18
using namespace std;
int n,f[maxn],s[maxn];
LL a,b,x[maxn];
int Get(int l,int r)
{
	if(l>r) return 0;
	if(!l) return s[r];
	else return (s[r]-s[l-1]+mod)%mod;
}
int main()
{
	scanf("%d%lld%lld",&n,&a,&b);
	if(a<b) swap(a,b);
	for(int i=1;i<=n;i++) scanf("%lld",&x[i]);
	x[0]=-INF,x[n+1]=INF;
	f[0]=s[0]=1;
	int l=0,r=0;
	for(int i=1;i<=n+1;i++)
	{
		while(r+1<i&&x[i]-x[r+1]>=a) r++;
		if(i>=2&&x[i-1]-x[i-2]<b) l=i-2;
		if(x[i]-x[i-1]>=a) f[i]=f[i-1];
		f[i]=(f[i]+Get(l,min(i-2,r)))%mod;
		s[i]=s[i-1];
		if(x[i+1]-x[i-1]>=b) s[i]=(s[i]+f[i])%mod;
	}
	printf("%d\n",f[n+1]);
}