Problem:
Score : 1100 points

Problem Statement

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.

Sample Input 1

5 3 7
1
3
6
9
12

Sample Output 1

5

There are five ways to perform division:

X={1,6,9,12}, Y={3}

X={1,6,9}, Y={3,12}

X={3,6,9,12}, Y={1}

X={3,6,9}, Y={1,12}

X={3,6,12}, Y={1,9}

Sample Input 2

7 5 3
0
2
4
7
8
11
15

Sample Output 2

4

Sample Input 3

8 2 9
3
4
5
13
15
22
26
32

Sample Output 3

13

Sample Input 4

3 3 4
5
6
7

Sample Output 4

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define F first
#define S second
#define pii pair<int,int>
typedef long long ll;
using namespace std;
const int N = 2 * 1e5 + 10 , mod = 1e9 + 7;
ll dp1[N] , dp2[N] , ps[N] , s[N];
int main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    int n , x , y; cin >> n >> x >> y; if(x > y)swap(x , y);
    for(int i=0;i<n;i++)cin >> s[i];
    for(int i=1;i<n;i++)ps[i] = ps[i - 1] + (s[i] - s[i - 1] < x);
    dp1[0] = dp2[0] = 1;
    dp1[1] = dp2[1] = 1;
    for(int i=2;i<=n;i++){
        //dp1[i]:
        if(s[i - 1] - s[i - 2] >= x)dp1[i] += dp1[i - 1];
        if(i == 2 || s[i - 1] - s[i - 3] >= x)dp1[i] += dp2[i - 1];
        dp1[i] %= mod;
        if(i > 2 && s[i - 1] - s[i - 3] < x)return cout << 0 , 0;
        //dp2[i];
        if(s[i - 1] - s[i - 2] >= y)dp2[i] += dp2[i - 1];
        int indr = upper_bound(s , s + i - 1 , s[i - 1] - y) - s;
        //indr : i - 1;
        if(indr != i - 1 && ps[i - 2] - ps[indr] == 0)dp2[i] += dp1[indr + 1];
        else if(indr == i - 1)dp2[i] += dp1[indr];
        dp2[i] %= mod;
    }
    cout << (dp1[n] + dp2[n]) % mod;
}