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;
typedef long long ll;
#define M 1000000007
#define INF 1000000000000000000
struct SEG{
  ll seg[1<<18];
  SEG(void){for(int i=0;i<1<<18;i++)seg[i]=0;}
  void add(int a,ll x){
      a+=(1<<17)-1;
      seg[a]=(seg[a]+x)%M;
      while(a>0){
          a=(a-1)/2;
          seg[a]=(seg[a*2+1]+seg[a*2+2])%M;
      }
  }
  ll sum(int a,int b,int l,int r,int k){
      if(r<a||b<l)return 0;
      if(a<=l&&r<=b)return seg[k];
      return (sum(a,b,l,(l+r-1)/2,k*2+1)+sum(a,b,(l+r+1)/2,r,k*2+2))%M;
  }
};
ll n,A,B,ae,be,s[100005];
SEG a,b;
int main(void){
    scanf("%lld%lld%lld",&n,&A,&B);
    s[0]=-INF;
    for(int i=1;i<=n;i++)scanf("%lld",s+i);
    a.add(0,1);
    b.add(0,1);
    for(int i=2;i<=n;i++){
        int p=upper_bound(s,s+n+1,s[i]-A)-s;
        if(p-1>=0&&be<=p-1)a.add(i-1,b.sum(be,p-1,0,(1<<17)-1,0));
        p=upper_bound(s,s+n+1,s[i]-B)-s;
        p=min(p,i-1);
        if(p-1>=0&&ae<=p-1)b.add(i-1,a.sum(ae,p-1,0,(1<<17)-1,0));
        if(s[i]<s[i-1]+A)ae=i-1;
        if(s[i]<s[i-1]+B)be=i-1;
    }
    printf("%lld\n",(a.sum(ae,n,0,(1<<17)-1,0)+b.sum(be,n,0,(1<<17)-1,0))%M);
}