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 SZ 1234567
typedef long long ll;
int n,g=0; ll A,B,s[SZ],x[SZ];
const int MOD=1e9+7;
struct Bit{
int ts,b[SZ],t[SZ]; void clr() {++ts;}
Bit() {ts=0; memset(t,0,sizeof t);}
void edt(int x,int y)
{
	for(;x<=g;x+=x&-x)
	{if(t[x]!=ts) t[x]=ts,b[x]=0; (b[x]+=y)%=MOD;}
}
int sum(int x)
{
	int s=0;
	for(;x>=1;x-=x&-x)
	{if(t[x]!=ts) t[x]=ts,b[x]=0; (s+=b[x])%=MOD;}
	return s;
}
ll ss(ll p)
{return sum(upper_bound(x+1,x+1+g,p)-x-1);}
ll ss(ll l,ll r)
{
	if(l<=r)
		return (ss(r)-ss(l-1))%MOD;
	return 0;
}
}ba,bb;
#define FO(x) {freopen(#x".in","r",stdin);freopen(#x".out","w",stdout);}
int main()
{
	scanf("%d%lld%lld",&n,&A,&B);
	for(int i=1;i<=n;++i)
		scanf("%lld",s+i),x[i]=s[i];
	g=n; x[++g]=-1e18;
	sort(x+1,x+1+g);
	ba.clr(); bb.clr();
	ba.edt(1,1); bb.edt(1,1);
	for(int i=2;i<=n;++i)
	{
		int r=lower_bound(x+1,x+1+g,s[i-1])-x;
		int s1=ba.sum(g)-ba.ss(s[i]-B+1,s[i]+B-1),
		s2=bb.sum(g)-bb.ss(s[i]-A+1,s[i]+A-1);
		s1%=MOD; s2%=MOD;
		if(abs(s[i]-s[i-1])<A) ba.clr();
		if(abs(s[i]-s[i-1])<B) bb.clr();
		ba.edt(r,s2); bb.edt(r,s1);
	}
	ll ans=ba.sum(g)+bb.sum(g);
	ans=(ans%MOD+MOD)%MOD;
	printf("%d\n",int(ans));
}