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 <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
ll f[100005],a[100005],A,B;
int n,pre[100005],sum[100005];
int main(){
	scanf("%d%lld%lld",&n,&A,&B);
	for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
	a[++n]=5e18;
	pre[1]=1;
	for(int i=2;i<n;i++){
		pre[i]=pre[i-1];
		while(a[i+1]-a[pre[i]]>=B&&pre[i]<i) pre[i]++;
	}
	for(int i=2;i<=n;i++) sum[i]=sum[i-1]+(a[i]-a[i-1]<A);
	for(int i=1;i<n;i++){
		if(sum[i]-sum[pre[i]]) pre[i]=n+1;
	}
	ll sum=1;f[0]=1;
	for(int i=1,L=1;i<=n;i++){
		if(i>2&&a[i-1]-a[i-2]<B) sum=0,L=i-2;
		while(a[i]-a[L]>=A&&L<i-1) sum=(sum+f[pre[L]])%mod,L++;
		f[i]=sum;
		if(i>1&&a[i]-a[i-1]>=A) f[i]=(f[i]+f[i-1])%mod;
	}
	printf("%lld",f[n]);
	return 0;
}