Problem: 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.

Examples

Input

5 3 7
1
3
6
9
12


Output

5


Input

7 5 3
0
2
4
7
8
11
15


Output

4


Input

8 2 9
3
4
5
13
15
22
26
32


Output

13


Input

3 3 4
5
6
7


Output

0

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
 
using namespace std;
 
 
typedef long long int64;
const int mod = 1e9 + 7;
 
struct BinaryIndexedTree
{
  vector< int > data;
 
  BinaryIndexedTree(int sz)
  {
    data.assign(++sz, 0);
  }
 
  int sum(int k)
  {
    int ret = 0;
    for(++k; k > 0; k -= k & -k) (ret += data[k]) %= mod;
    return (ret);
  }
 
  void add(int k, int x)
  {
    for(++k; k < data.size(); k += k & -k) (data[k] += x) %= mod;
  }
 
  int get(int k)
  {
    return ((sum(k) - sum(k - 1) + mod) % mod);
  }
};
 
int main()
{
  int64 N, A, B, X[100000];
  int64 AA[100000], BB[100000];
 
  cin >> N >> A >> B;
  for(int i = 0; i < N; i++) {
    cin >> X[i];
    AA[i] = upper_bound(X, X + i + 1, X[i] - A) - X;
    BB[i] = upper_bound(X, X + i + 1, X[i] - B) - X;
  }
 
  BinaryIndexedTree dp1(N), dp2(N);
  dp1.add(0, 1);
  dp2.add(0, 1);
  int aa = 0, bb = 0;
 
  for(int i = 1; i < N; i++) {
    int s = dp2.sum(AA[i]), t = dp1.sum(BB[i]);
    dp1.add(i, s);
    dp2.add(i, t);
    if(X[i] - X[i - 1] < A) 
      for(; aa < i; aa++) dp1.add(aa, mod - dp1.get(aa));
    if(X[i] - X[i - 1] < B)
      for(; bb < i; bb++) dp2.add(bb, mod - dp2.get(bb));
  }
 
  cout << (dp1.sum(N - 1) + dp2.sum(N - 1)) % mod << endl;
}