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;

int N;
ll mod = 1000000007;
ll A,B;
ll S[100010];
ll bit[100010];

void init()
{
    for(int i = 0; i <= N + 10; i++)bit[i] = 0;
    return;
}

void add(int i,ll x)
{
    i += 2;
    while(i <= N + 10)
    {
        bit[i] = (bit[i] + mod + x) % mod;
        i += i & -i;
    }
    return;
}

ll sum(int i)
{
    i += 2;
    ll res = 0;
    while(i > 0)
    {
        res = (res + bit[i]) % mod;
        i -= i & -i;
    }
    return res;
}

int main()
{
    scanf("%d%lld%lld",&N,&A,&B);
    for(int i = 1; i <= N; i++)scanf("%lld",S + i);
    S[0] = -1001001001001001001;

    if(A > B)swap(A,B);
    init();

    add(0,1LL);
    add(1,1LL);

    int z = 0,l = 0;
    for(int i = 2; i <= N; i++)
    {
        if(S[i] - S[i - 2] < A)

        {
            printf("0\n");
            return 0;
        }
        while(S[i] - S[l] >= B)l++;
        add(i,sum(l - 1));

        if(S[i] - S[i - 1] < A)
            for(;z < i - 1; z++)add(z,-sum(z));
    }
    printf("%lld\n",sum(N));
    return 0;
}