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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
int dp[2][200000 + 3];
int c[100000 + 1];

static inline int fg(int i, int j){
 return (i < j) ? i : i+j ;
}

int main() {
	int n;
  int a;
  int b;
  scanf("%d", &n);
  scanf("%d", &a);
  scanf("%d", &b);
  int min_diff = (a<b)? b+1 : a+1;
  scanf("%d", &c[0]);
  for(int i=1; i<n; i++){
    scanf("%d", &c[i]);
    if (c[i]-c[i-1]<min_diff) min_diff=c[i]-c[i-1];
    //min_diff = (c[i] - c[i-1] > min_diff) ? min_diff :c[i]-c[i-1];
  }
  //printf("%d\n", min_diff);

  dp[1][0]=1;
  dp[1][1]=1;

  for(int i=1; i<=n-2; i++){
    for(int j=0; j<=i-1; j++){
    int ij = dp[i%2][fg(i,j)]%1000000007;
    int ji = dp[i%2][fg(j,i)]%1000000007;
    if(c[i+1] - c[i] >= a) dp[(i+1)%2][fg(i+1,j)] += ij;// if c[i+1] - c[i] >= a
    if(c[i+1] - c[j] >= b) dp[(i+1)%2][fg(i,i+1)] += ij; 
    if(c[i+1] - c[i] >= b) dp[(i+1)%2][fg(j,i+1)] += ji; 
    if(c[i+1] - c[j] >= a) dp[(i+1)%2][fg(i+1,i)] += ji;
    dp[i%2][fg(i,j)]=0;
    dp[i%2][fg(j,i)]=0;
    }
  }

int sum = 0;
  if(min_diff>=a)sum += 1;
  if(min_diff>=b)sum += 1;
//sum += 1 if c.each_cons(2).map{|k| (k[1]-k[0]) >= a[1]}.all?#(TRUE)
//sum += 1 if c.each_cons(2).map{|k| (k[1]-k[0]) >= a[2]}.all?#(TRUE)
  for(int j=0;j<2*n-1;j++){
    sum += dp[(n-1)%2][j] % 1000000007;
  //printf("%d %d\n",j, dp[(n-1)%2][j]);
  //sum += dp[(n-1)%2][j];
  }
  printf("%d\n", sum);
  
}