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<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f7f7f7f7f
#define ll long long
using namespace std;
const int mod=1e9+7;
const int maxn=1e5;
int n;
int f[maxn+8],g[maxn+8];
ll A,B;
ll a[maxn+8];

ll read()
{
    ll x=0,f=1;char ch=getchar();
    for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
    for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
    return x*f;
}

int main()
{
    n=read(),A=read(),B=read();
    for (int i=1;i<=n;i++) a[i]=read();a[0]=-inf,a[n+1]=inf;
    f[0]=g[0]=1;
    int j_a=0,j_b=0,pre_a=0,pre_b=0;
    for (int i=1;i<=n;i++)
	{
	    //printf("Time:%d\n",i);
	    if (a[i]-a[i-1]<A) pre_a=i-1;
	    if (a[i]-a[i-1]<B) pre_b=i-1;
	    j_a++;while(j_a<i&&a[i+1]-a[j_a]>=A) j_a++;j_a--;
	    j_b++;while(j_b<i&&a[i+1]-a[j_b]>=B) j_b++;j_b--;
	    //printf("%d %d\n",pre_a,pre_b);
	    if (j_a>=pre_b) g[i]=pre_b?(f[j_a]-f[pre_b-1])%mod:f[j_a];
	    if (j_b>=pre_a) f[i]=pre_a?(g[j_b]-g[pre_a-1])%mod:g[j_b];
	    //printf("check:%d %d\n",f[i],g[i]);
	    f[i]=(f[i]+f[i-1])%mod;
	    g[i]=(g[i]+g[i-1])%mod;
	}
    int ans=((((f[n]-f[n-1])%mod+g[n])%mod-g[n-1])%mod+mod)%mod;
    printf("%d\n",ans);
    return 0;
}