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 LL long long
using namespace std;

const int N = 100000+9;
const int MOD = 1000000007;
const int INF = 4e18;

LL A,B,LIM[2],s[N];
int n,f[2][N],sum[2][N],lst[2][N],pre[2][N];

inline LL read() {
	char c=getchar(); LL f=1,ret=0;
	while (c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
	while (c<='9'&&c>='0') {ret=ret*10+c-'0';c=getchar();}
	return ret * f;
}

int main() {
	//freopen("11input.in","r",stdin);
	n = read(); A = LIM[0] = read(); B = LIM[1] = read();
	for (int i=1,l1=1,l2=1,p1=1,p2=1;i<=n;i++) {
		s[i] = read();
		while (l1 < i && s[i] - s[l1+1] >= A) l1++; lst[0][i] = l1;
		while (l2 < i && s[i] - s[l2+1] >= B) l2++; lst[1][i] = l2;
		if (i > 1 && s[i] - s[i-1] < A) p1 = i - 1; pre[0][i] = p1;
		if (i > 1 && s[i] - s[i-1] < B) p2 = i - 1; pre[1][i] = p2; 
	}
	lst[0][n+1] = lst[1][n+1] = n;
	s[n+1] = INF; sum[1][1] = sum[0][1] = 1;
	for (int i=2,l,r;i<=n;i++) {
		for (int j=0;j<=1;j++) {
			r = min(i-1, lst[j^1][i+1]);
			if (s[i+1] - s[lst[j^1][i+1]] >= LIM[j^1]) 
				if (l = pre[j][i], l <= r) f[j][i] = (sum[j^1][r] - sum[j^1][l-1]) % MOD;
			sum[j][i] = (sum[j][i-1] + f[j][i]) % MOD;
		}
	}
	printf("%d\n",((f[0][n]+f[1][n])%MOD+MOD)%MOD);
	return 0;
}