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 <bits/stdc++.h>

using namespace std;

//#define FILE_IO

typedef long long LL;

const LL INF = 1LL << 61;
const int mod = 1e9 + 7;

int N;
LL A, B;
LL v[100005];
int dp[100005][2];

int add(int x, int y) {x += y; if(x >= mod) x -= mod; return x;}
int sub(int x, int y) {x -= y; if(x < 0) x += mod; return x;}

int main()
{
    #ifdef FILE_IO
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    #endif

    scanf("%d", &N);
    scanf("%lld%lld", &A, &B);
    if(A > B)   swap(A, B);
    v[0] = -INF;
    for(int i = 1; i <= N; i++) scanf("%lld", &v[i]);

    dp[1][0] = dp[1][1] = 1;
    int posB = 1;
    int lftA = 1;
    for(int i = 2; i <= N; i++)
    {
        if(v[i] - v[i - 1] >= A)    dp[i][0] = add(dp[i - 1][0], dp[i - 1][1]);
        else if(v[i] - v[i - 2] >= A)   dp[i][0] = dp[i - 1][1];
        else { if(i > 2) { printf("0\n"); return 0; } }

        while(v[i] - v[posB] >= B)  posB++;

        if(posB == i)   dp[i][1] = add(dp[i - 1][0], dp[i - 1][1]);
        else if(lftA <= posB)   dp[i][1] = dp[posB][0];

        if(v[i] - v[i - 1] >= A);
        else    lftA = i;
    }

    int ans = add(dp[N][0], dp[N][1]);
    printf("%d\n", ans);

    return 0;
}