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>
#include <cassert>
#include <cctype>
#include <algorithm>
using namespace std;
typedef long long lint;
#define cout cerr
#define ni (next_num<int>())
#define nl (next_num<lint>())
template<class T>inline T next_num(){
	T i=0;char c;
	while(!isdigit(c=getchar())&&c!='-');
	bool flag=c=='-';
	flag?(c=getchar()):0;
	while(i=i*10-'0'+c,isdigit(c=getchar()));
	return flag?-i:i;
}
const int N=100010,O=1000000007;
inline int add(const int &a,const int &b){return (a+b)%O;}
inline int sub(const int &a,const int &b){return add(a,O-b);}
const lint LINF=0x7f7f7f7f7f7f7f7f;
lint a[N];
int f[N],g[N],_f[N],_g[N];
int main(){
	int n=ni;
	lint da=nl,db=nl;
	for(int i=1;i<=n;i++){
		a[i]=nl;
	}
	a[0]=-LINF,a[n+1]=LINF;
	sort(a+1,a+n+1);
	f[0]=g[0]=_f[0]=_g[0]=1;
	int al=1,bl=1,ar=0,br=0;
	for(int i=1;i<=n;i++){
		if(a[i-1]+da>a[i]){
			al=i;
		}
		if(a[i-1]+db>a[i]){
			bl=i;
		}
		for(;ar<i&&a[ar]+db<=a[i+1];ar++);
		for(;br<i&&a[br]+da<=a[i+1];br++);
		f[i]=al<=ar?sub(_g[ar-1],al>=2?_g[al-2]:0):0;
		g[i]=bl<=br?sub(_f[br-1],bl>=2?_f[bl-2]:0):0;
		_f[i]=(_f[i-1]+f[i])%O;
		_g[i]=(_g[i-1]+g[i])%O;
	}
	printf("%d\n",(f[n]+g[n])%O);
	return 0;
}