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 <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

typedef long long i64;
const int MAX_N = 100005, MOD = 1000000007;
const i64 INF = 0x3f3f3f3f3f3f3f3fLL;

i64 f[MAX_N], g[MAX_N];
int mark1[MAX_N], mark2[MAX_N];

int N;
i64 A, B, S[MAX_N];

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

  int order = 0, rm1 = 0, rm2 = 0;
  i64 sum1 = 0, sum2 = 0;
  
  f[0] = g[0] = 1;

  int l1 = 0, l2 = 0;

  S[0] = -INF;
  for (int i = 2; i <= N; ++i) {
    ++order;
    while (l1 < i - 1 && S[i] - S[l1] >= A) {
      if (mark2[l1] >= rm2) sum1 = (sum1 + g[l1]) % MOD;
      l1++;
    }
    
    while (l2 < i - 1 && S[i] - S[l2] >= B) {
      if (mark1[l2] >= rm1) sum2 = (sum2 + f[l2]) % MOD;
      l2++;
    }
    
    f[i - 1] = (f[i - 1] + sum1) % MOD, mark1[i - 1] = order;
    g[i - 1] = (g[i - 1] + sum2) % MOD, mark2[i - 1] = order;
    if (S[i] - S[i - 1] < A) rm1 = order, sum2 = 0;
    if (S[i] - S[i - 1] < B) rm2 = order, sum1 = 0;
  }

  i64 res = 0;
  for (int i = 0; i < N; ++i) {
    if (mark1[i] >= rm1) res = (res + f[i]) % MOD;
    if (mark2[i] >= rm2) res = (res + g[i]) % MOD;
  }

  printf("%lld\n", res);
  return 0;
}