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 Allah
#include <bits/stdc++.h>
using namespace std;

#define set SET
const int maxN = 100 * 1000 + 10; 
typedef long long ll; 
const int mod = 1e9 + 7; 

inline int mul( int a , int b ) { 
    return a * 1ll * b  % mod; 
}
inline int sum( int a , int b ) { 
    int c = a + b; 
    if( c >= mod ) c -= mod;
    return c;
}
int n;

ll s[maxN];
int dp[2][maxN];

int main() { 
    cin >> n;
    ll a[2]; cin >> a[0] >> a[1]; 


    n++;
    s[0] = 1e9;
    s[0] = -(s[0] * s[0] + 1000 );
    cin >> s[1]; 
    dp[0][0] = dp[1][0] = 1; 

    int st[2] = { 0 , 0 }; 

    for( int i = 2 ; i < n ; i++ ) { 
        cin >> s[i]; 
        for( int t = 0 ; t < 2 ; t++ ) { 
            int v = upper_bound( s , s + i - 1 , s[i] - a[t] ) - s;
            dp[t][i-1] = sum(dp[t][i-2] , dp[t^1][v-1]);
        }
        for( int t = 0 ; t < 2 ; t++ ) 
            if( s[i] - s[i-1] < a[t] ) { 
                dp[t][i-1] = sum( dp[t][i-1] , mod - dp[t][i-2] ); 
                for( ;st[t] < i - 1 ; st[t]++ ) 
                    dp[t][st[t]] = 0; 
            }
    }
    cout << sum(dp[0][n-2],dp[1][n-2]) << endl;
}