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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll mod = 1000000007;
#define SIZE 131072
class BIT
{
public:
	ll bit[SIZE + 1];
	void add(int a, ll b)
	{
		a+=5;
		for (;;)
		{
			bit[a] += b;
			bit[a] %= mod;
			a += a&-a;
			if (a > SIZE)return;
		}
	}
	ll get(int a)
	{
		a+=5;
		ll ret = 0;
		for (;;)
		{
			ret += bit[a];
			ret %= mod;
			a -= a&-a;
			if (a == 0)return ret;
		}
	}
};
BIT bi;
int main()
{
	int num;
	ll da, db;
	scanf("%d%lld%lld", &num, &da, &db);
	if (da > db)swap(da, db);
	vector<ll>vec;
	vec.push_back(-3000000000000000000LL);
	for (int i = 0; i < num; i++)
	{
		ll z;
		scanf("%lld", &z);
		vec.push_back(z);
	}
	bi.add(0, 1);
	int pt = 0;
	for (int i = 0; i < num; i++)
	{
		int upp = upper_bound(vec.begin(), vec.end(), vec[i + 1] - db) - vec.begin() - 1;
		ll t;
		if (upp >= pt)t = bi.get(upp) - bi.get(pt - 1);
		else t = 0;
		if (vec[i + 1] - vec[i] < da)pt = i;
		if (i > 0)if (vec[i + 1] - vec[i - 1] < da)pt = i + 1;
		bi.add(i + 1, (t + mod) % mod);
	}
	if (num >= pt)printf("%lld\n", (bi.get(num) - bi.get(pt - 1) + mod) % mod);
	else printf("0\n");
}