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:
//In the name of God
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;

const ll maxn = 2e5 + 100;
const ll mod = 1e9 + 7;
const ll inf = 1e18;

#define fast_io ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define file_io freopen("input.txt", "r+", stdin);freopen("output.txt", "w+", stdout);
#define pb push_back
#define Mp make_pair
#define F first
#define S second

ll n, dp[maxn][2], a[maxn], x[2], d[2];

int main(){
    fast_io;

    cin >> n >> x[0] >> x[1];
    a[0] = -inf;
    for(ll i = 1; i <= n; i++){
        cin >> a[i];
    }
    dp[1][0] = 1;
    dp[1][1] = 1;
    for(ll i = 2; i <= n; i++){
        for(ll j = 0; j < 2; j++){
            if(a[i] - a[i - 1] >= x[j]){
                dp[i][j] = (dp[i - 1][j] + dp[i - 1][1 - j]) % mod;
            }
            else{
                ll y = upper_bound(a, a + i, a[i] - x[j]) - a;
                if(d[1 - j] < y){
                    dp[i][j] = dp[y][1 - j];
                }
            }
        }
        if(a[i] - a[i - 1] < x[0]){
            d[0] = i - 1;
        }
        if(a[i] - a[i - 1] < x[1]){
            d[1] = i - 1;
        }
    }
    cout << (dp[n][0] + dp[n][1]) % mod;

    return 0;
}