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;
void read(int& x){ scanf("%d",&x); }
void read(ll& x){ scanf("%lld",&x); }
template<typename T,typename... Args>
void read(T& a,Args&... b){ read(a); read(b...); }

int n;
ll a, b;
ll s[100010];

const ll mod = ll(1e9) + 7;
const ll inf = ll(1e18) + 10;

ll ca[100010];
ll cb[100010];
ll pca[100010];
ll pcb[100010];

int iha(ll x){
    return upper_bound(s, s+n+1, x) - s - 1;
}

int main()
{
    read(n, a, b);
    for(int i=1; i<=n; ++i) read(s[i]);
    s[0]=-inf;
    
    int za=-1, zb=-1;
    
    auto f=[&](ll* arr,int zero,int x){
        if(x < zero) return 0LL;
        ll ret=arr[x] + mod;
        if(zero >= 0) ret -= arr[zero];
        return ret%mod;
    };
    
    ca[0]=pca[0]=1;
    cb[0]=pcb[0]=1;
    
    for(int i=2; i<=n; ++i){
        int ia=min(iha(s[i]-a), i-2);
        int ib=min(iha(s[i]-b), i-2);
        cb[i-1]=f(pca, za, ia);
        ca[i-1]=f(pcb, zb, ib);
        pca[i-1]=(pca[i-2]+ca[i-1])%mod;
        pcb[i-1]=(pcb[i-2]+cb[i-1])%mod;
        if(s[i]-s[i-1]<a) zb=i-2;
        if(s[i]-s[i-1]<b) za=i-2;
    }
    
    printf("%lld\n", (f(pca, za, n-1)+f(pcb, zb, n-1))%mod);
    return 0;
}