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<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int N=100005;
typedef long long ll;
const ll mod=1e9+7;
inline void MOD(int &x) {if (x>=mod) x-=mod;}


int n;
ll S[N],A,B;
int dp[N][2],s[N][2];
int W(int id,int l,int r)
{
	if (l>r) return 0;
	if (!l) return s[r][id]+1;
	return (s[r][id]-s[l-1][id]+mod)%mod;
}
int main()
{
	scanf("%d",&n);
	scanf("%lld%lld",&A,&B);
	for (int i=1;i<=n;i++) scanf("%lld",&S[i]);
	
	s[1][0]=s[1][1]=dp[1][0]=dp[1][1]=1;
	int la=0,lb=0;
	int pa=0,pb=0;
	for (int i=2;i<=n-1;i++)
	{
		while (S[pa+1]+A<=S[i+1]&&pa+1<i) pa++;
		while (S[pb+1]+B<=S[i+1]&&pb+1<i) pb++;
		if (S[i]-S[i-1]<A) la=i-1;
		if (S[i]-S[i-1]<B) lb=i-1;
		
		dp[i][0]=W(1,la,pb);MOD(s[i][0]=s[i-1][0]+dp[i][0]);
		dp[i][1]=W(0,lb,pa);MOD(s[i][1]=s[i-1][1]+dp[i][1]);
	/*	printf("i=%d la=%d lb=%d pa=%d pb=%d dp0=%d dp1=%d\n"
		,i,la,lb,pa,pb,dp[i][0],dp[i][1]);*/
	}
	
	
	
	dp[0][0]=dp[0][1]=1;

	int ans=dp[n-1][0];
	for (int i=n-1;i>=1;i--) 
	{
		if (S[i+1]-S[i]<B) break;
		MOD(ans+=dp[i-1][0]);
	}
	MOD(ans+=dp[n-1][1]);
	for (int i=n-1;i>=1;i--) 
	{
		if (S[i+1]-S[i]<A) break;
		MOD(ans+=dp[i-1][1]);
	}
	printf("%d\n",ans);
	return 0;
}
/*
5 3 7
1 3 6 9 12
*/