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>
using namespace std;

const int mod = 1e9+7;
int n;
long long a, b, s[100001];

struct Seg{
	int *data, base;
	Seg(int sz){
		for(base=1; base<sz; base*=2);
		data=new int[base*2];
	}
	void update(int p, int v){
		p++;
		p+=base-1;
		data[p]=v;
		for(p/=2; p; p/=2)
			data[p]=(data[p*2]+data[p*2+1])%mod;
	}
	int rangeSum(int s, int e){
		s++, e++;
		int ret=0;
		for(s+=base-1, e+=base-1; s<=e; s/=2, e/=2){
			if(s&1) ret=(ret+data[s++])%mod;
			if(!(e&1)) ret=(ret+data[e--])%mod;
		}
		return ret;
	}
};

int get(int l, int r, long long val){
	if(l>r) return -1e9;
	int mid=(l+r)/2;
	if(mid==0 || s[mid]<=val) return max(mid, get(mid+1, r, val));
	else return get(l, mid-1, val);
}

int main(){
	scanf("%d %lld %lld", &n, &a, &b);
	for(int i=1; i<=n; i++)
		scanf("%lld", &s[i]);

	Seg *x=new Seg(n), *y=new Seg(n);
	int sx=0, sy=0;

	x->update(0, 1), y->update(0, 1);
	for(int i=1; i<n; i++){
		y->update(i, x->rangeSum(sx, get(sx, i-1, s[i+1]-b)));
		x->update(i, y->rangeSum(sy, get(sy, i-1, s[i+1]-a)));
		if(s[i+1]-s[i] < a) sx=i;
		if(s[i+1]-s[i] < b) sy=i;
	}

	int ans = (x->rangeSum(sx, n-1) + y->rangeSum(sy, n-1))%mod;
	printf("%d\n", ans);
	return 0;
}