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 <algorithm>
using namespace std;
struct segment_tree
{
	int seg[262144];
	int query(int id)
	{
		int ret=0;
		for (id+=131072;id;id>>=1)
			(ret+=seg[id])%=1000000007;
		return ret;
	}
	void add(int l,int r,int val)
	{
		for (l+=131072,r+=131072;l<r;l>>=1,r>>=1)
		{
			if (l&1)
				(seg[l++]+=val)%=1000000007;
			if (r&1)
				(seg[--r]+=val)%=1000000007;
		}
	}
};
segment_tree seg1,seg2;
long long s[100001];
int nxt1[100001],nxt2[100001];
int main()
{
	int n;
	long long a,b;
	scanf("%d%lld%lld",&n,&a,&b);
	for (int i=1;i<=n;++i)
		scanf("%lld",s+i);
	nxt1[n]=nxt2[n]=n;
	for (int i=n-1;i>=1;--i)
	{
		nxt1[i]=(s[i+1]-s[i]<a?i:nxt1[i+1]);
		nxt2[i]=(s[i+1]-s[i]<b?i:nxt2[i+1]);
	}
	seg1.add(0,1,1);
	seg2.add(0,1,1);
	for (int i=1;i<=n;++i)
	{
		int val1=seg1.query(i-1),val2=seg2.query(i-1);
		if (val1)
		{
			int nxt=nxt2[i],fir1=(i==1?2:lower_bound(s+i,s+n+1,s[i-1]+a)-s);
			seg2.add(fir1-1,nxt+1,val1);
		}
		if (val2)
		{
			int nxt=nxt1[i],fir2=(i==1?2:lower_bound(s+i,s+n+1,s[i-1]+b)-s);
			seg1.add(fir2-1,nxt+1,val2);
		}
	}
	printf("%d",(seg1.query(n)+seg2.query(n))%1000000007);
	return 0;
}