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;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

const int mod=1000000007;

const int INF=1001001001001001001ll*3;

int N,A,B;

int S[111111];

int sum[111111];
int dp[111111];

signed main(){
    scanf("%lld%lld%lld",&N,&A,&B);
    if(A>B)swap(A,B);

    S[0]=-INF;
    rep(i,N)scanf("%lld",&S[i+1]);
    S[N+1]=INF;

    N+=2;

    rep(i,N-2){
        if(S[i+2]-S[i]<A){
            cout<<0<<endl;
            return 0;
        }
    }

    dp[0]=1;
    sum[0]=1;

    int l=0;
    int r=0;
    for(int i=1;i<N;i++){
        while(S[i]-S[r+1]>=B)r++;
        if(l<=r){
            dp[i]=sum[r];
            if(l)dp[i]=(dp[i]-sum[l-1]+mod)%mod;
        }
        sum[i]=(sum[i-1]+dp[i])%mod;

        if(S[i]-S[i-1]<A)l=i-1;
    }

    cout<<dp[N-1]<<endl;
    return 0;
}