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: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll mod = 1e9+7, inf = 2e18;
ll n, a, b, x[100005], za, zb;
vector<ll> ca, cb;

ll getA (ll X) {
	if(X < za) return 0;
	if(za) return (ca[X] - ca[za-1] + mod) % mod;
	return ca[X];
}

ll getB (ll X) {
	if(X < zb) return 0;
	if(zb) return (cb[X] - cb[zb-1] + mod) % mod;
	return cb[X];
}

int main()
{
	scanf("%lld%lld%lld",&n,&a,&b);
	for(int i=1;i<=n;i++) {
		scanf("%lld",&x[i]);
	}
	x[0] = -inf;
	ca.push_back(1);
	cb.push_back(1);
	for(int i=2;i<=n;i++) {
		ll fa = min(upper_bound(x, x+n+1, x[i]-a) - x - 1ll, i - 2ll);
		ll fb = min(upper_bound(x, x+n+1, x[i]-b) - x - 1ll, i - 2ll);
		ca.push_back((getB(fb)+ca.back())%mod);
		cb.push_back((getA(fa)+cb.back())%mod);
		if(x[i-1] + a > x[i]) zb = i-1;
		if(x[i-1] + b > x[i]) za = i-1;
	}
	printf("%lld\n",(getA(n-1)+getB(n-1)) % mod);
}