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 <iostream>
#include <algorithm>
#include <cassert>
using namespace std;

const int MAX_N = 1e5;
const long long MOD = 1e9 + 7;
int N;
long long A, B;
long long S[MAX_N + 1];

long long dp_x[MAX_N] = {};
long long dp_y[MAX_N] = {};
int x_last; // X の条件を満たす最大添え字+1
int y_last; // Y の条件を満たす最大添え字+1
int x_continue; // X のみで構成できる添え字の開始位置
int y_continue; // Y のみで構成できる添え字の開始位置

long long solve()
{
	dp_x[0] = 1;
	dp_y[0] = 1;
	x_last = 0;
	y_last = 0;
	x_continue = 0;
	y_continue = 0;

	for (int i = 1; i < N; ++i)
	{
		while (S[i] - S[x_last] >= A) ++x_last;
		while (S[i] - S[y_last] >= B) ++y_last;

		if (S[i] - S[i - 1] >= A)
			dp_x[i] = dp_x[i - 1];
		if (x_last >= y_continue)
			dp_x[i] = (dp_x[i] + dp_y[min(x_last, i-1)]) % MOD;

		if (S[i] - S[i - 1] >= B)
			dp_y[i] = dp_y[i - 1];
		if (y_last >= x_continue)
			dp_y[i] = (dp_y[i] + dp_x[min(y_last, i-1)]) % MOD;

		if (S[i] - S[i - 1] < A)
			x_continue = i;
		if (S[i] - S[i - 1] < B)
			y_continue = i;
	}

	return (dp_x[N - 1] + dp_y[N - 1]) % MOD;
}

int main()
{
	cin >> N >> A >> B;
	for (int i = 0; i < N; ++i)
	{
		cin >> S[i];
	}
	cout << solve() << endl;
}