p03823 AtCoder Grand Contest 009 - Division into Two
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
```python
import sys
input = sys.stdin.readline

from bisect import bisect_left, bisect_right

INF = 10**18 + 100
N,A,B = map(int,input().split())
S = [-INF] + [int(x) for x in sys.stdin.read().split()]

MOD = 10**9 + 7

dpX = [0] * (N+1) # 最後にYを選んだとして、直前に選んだXがどこにあるか
dpY = [0] * (N+1) # 最後にXを選んだとして、直前に選んだYがどこにあるか
dpX[0] = 1
dpY[0] = 1
dpX_cum = [1] * (N+1) + [0] # -1
dpY_cum = [1] * (N+1) + [0] # -1
dpX_left = 0
dpY_left = 0
for n,x in enumerate(S[2:],2):
    iA = bisect_right(S,x-A)
    iB = bisect_right(S,x-B)
    # ....XY
    xy = dpY_cum[iB-1] - dpY_cum[dpY_left-1] if iB >= dpY_left else 0
    # ....YX
    yx = dpX_cum[iA-1] - dpX_cum[dpX_left-1] if iA >= dpX_left else 0
    # ....XX が不可能なら捨てる。明示的に捨てるのではなく、生きている番号だけ持つ
    if iA != n:
        dpY_left = n-1
    if iB != n:
        dpX_left = n-1
    dpX[n-1] = xy
    dpX_cum[n-1] = (dpX_cum[n-2] + xy) % MOD
    dpX_cum[n] = dpX_cum[n-1]
    dpY[n-1] = yx
    dpY_cum[n-1] = (dpY_cum[n-2] + yx) % MOD
    dpY_cum[n] = dpY_cum[n-1]

answer = dpX_cum[N-1] - dpX_cum[dpX_left-1]
answer += dpY_cum[N-1] - dpY_cum[dpY_left-1]
answer %= MOD
print(answer)
```