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 <iostream>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <vector>

using namespace std;

#define pb push_back
#define mp make_pair
#define ll long long
#define ull unsigned ll
#define db double
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define PII pair<int, int>

ll n,a,b;
ll dp[100010][2];
ll s[100010];
int len[100010];

int main() {
    scanf("%lld%lld%lld",&n,&a,&b);
    if (a>b) swap(a,b);
    for (int i=1;i<=n;i++) {
        scanf("%lld",s+i);
    }
    s[0]=-1e18;
    dp[1][0]=dp[1][1]=1;
    for (int i=2;i<=n;i++) {
        if (s[i]-s[i-1]>=a) len[i]=len[i-1]+1;
        if (s[i]-s[i-2]>=a) dp[i][0]=(dp[i][0]+dp[i-1][1])%MOD;
        if (s[i]-s[i-1]>=a) dp[i][0]=(dp[i][0]+dp[i-1][0])%MOD;
        if (s[i]-s[i-1]>=b) dp[i][1]=(dp[i-1][0]+dp[i-1][1])%MOD;
        else if (s[i]-s[i-2]>=b) dp[i][1]=dp[i-1][0]%MOD;
        else {
            int pos=upper_bound(s,s+i,s[i]-b)-s;
            if (i-1-len[i-1]<=pos) dp[i][1]=dp[pos][0];
        }
    }
    ll ans=(dp[n][0]+dp[n][1])%MOD;
    printf("%lld\n",ans);
}