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>
#define N 100100
#define Q 1000000007
using namespace std;
long long s[N];
long long a,b;
int nb[N],ca[N];
bool chk[N][2];
int dp[N][2];
int solve(int k,int st,int n){
	if(k>=n) return 1;
	if(chk[k][st]) return dp[k][st];
	chk[k][st]=true;
	if(st==0){
		if(s[k+1]-s[k]>=a) dp[k][st]=(dp[k][st]+solve(k+1,0,n))%Q;
		else dp[k][st]=(dp[k][st]+solve(k+2,1,n))%Q;
		dp[k][st]=(dp[k][st]+solve(k+1,1,n))%Q;
		return dp[k][st];
	}
	else{
		if(s[k]-s[k-1]>=b) return dp[k][st]=solve(k,0,n);
		if(k>=2&&s[k]-s[k-2]<a) return 0;
		if(ca[k]<nb[k-1]) return 0;
		else if(ca[k]==nb[k-1]) return dp[k][st]=solve(nb[k-1]+1,1,n);
		else return dp[k][st]=solve(nb[k-1],0,n);
	}
}
int main(){
	int n;
	scanf("%d%lld%lld",&n,&a,&b);
	if(a>b) swap(a,b);
	for(int i=0;i<n;i++) scanf("%lld",&s[i]);
	s[n]=1LL<<62;
	ca[n]=n+2,nb[n]=n;
	for(int i=n-1;i>=0;i--){
		if(s[i]+a<=s[i+1]) ca[i]=ca[i+1];
		else ca[i]=i+1;
		nb[i]=nb[i+1];
		while(s[nb[i]-1]-s[i]>=b) nb[i]--;
	}
	printf("%d\n",solve(0,0,n));
	return 0;
}