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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
 	s = 0; int f = 0; char c = getchar();
 	while (!isdigit(c)) f |= (c == '-'), c = getchar();
 	while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
 	s = f ? -s : s;
}

typedef long long LL;
const int _ = 100010;
const LL p = 1e9 + 7;

int n; LL dp[_], sum[_], A, B, a[_];

int main() {
	read(n), read(A), read(B); if (A < B) swap(A, B);
	for (rg int i = 1; i <= n; ++i) read(a[i]);
	for (rg int i = 1; i + 2 <= n; ++i)
		if (a[i + 2] - a[i] < B) return puts("0"), 0;
	dp[0] = sum[0] = 1;
	for (rg int l = 0, r = 0, i = 1; i <= n; ++i) {
		while (r < i && a[i] - a[r + 1] >= A) ++r;
		if (l <= r) dp[i] = (sum[r] - (l > 0 ? sum[l - 1] : 0ll) + p) % p;
		sum[i] = (sum[i - 1] + dp[i]) % p;
		if (i > 1 && a[i] - a[i - 1] < B) l = i - 1;
	}
	LL ans = 0;
	for (rg int i = n; ~i; --i) {
		ans = (ans + dp[i]) % p;
		if (i < n && a[i + 1] - a[i] < B) break;
	}
	printf("%lld\n", ans);
	return 0;
}