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)

#pragma optimization_level 3
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;

#define MOD 1000000007

struct BIT {
  int n;
  vector<int> xs;

  BIT(int n) : n(n) {
    xs.resize(n+1);
  }

  int sum(int i) {
    int s = 0;
    for (int x=i+1; x>0; x-=x&-x) s = (s + xs[x]) % MOD;
    return s;
  }

  int sum(int a, int b) {
    if (a > b) return 0;
    return (MOD + (long long)sum(b) - (long long)sum(a-1)) % MOD;
  }

  void add(int i, int v) {
    for (int x=i+1; x<=n; x+=x&-x) xs[x] = (xs[x] + v) % MOD;
  }
};

int N, A, B;
long long S[100001];

signed main() {
  cin >> N >> A >> B;
  if (A > B) swap(A, B);
  S[0] = -(1LL<<60);
  for (int i=1; i<=N; i++) cin >> S[i];

  BIT dp0(N+1), dp1(N+1);
  dp0.add(0, 1);
  dp1.add(0, 1);

  int l0 = 0, l1 = 0;
  for (int i=1; i<N; i++) {
    int uj1 = min(i, (int)(upper_bound(S, S+N+1, S[i+1]-A) - S) - 1);
    int uj2 = min(i, (int)(upper_bound(S, S+N+1, S[i+1]-B) - S) - 1);
    long long s1 = dp1.sum(l1, uj1), s2 = dp0.sum(l0, uj2);
    if (S[i+1]-S[i] < A) l0 = i;
    if (S[i+1]-S[i] < B) l1 = i;
    dp0.add(i, s1);
    dp1.add(i, s2);
  }

  int s = (dp0.sum(l0, N) + dp1.sum(l1, N)) % MOD;
  cout << s << "\n";
  return 0;
}