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 <vector>
#include <algorithm>
using namespace std;

const int MOD = 1000000007;

long long a, b;
long long s[100100];
long long dp[100100];
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;

long long sum[100100];
long long get(int l, int r) { //[l, r)
        if (l > r) return 0;
        if (l == 0) return sum[r];
        return sum[r] - sum[l - 1];
}

int main() {
        int n;
        scanf("%d", &n);
        scanf("%lld%lld", &a, &b);
        if (a > b) swap(a, b);
        s[0] = -INFL;
        for (int i = 1; i <= n; i ++) scanf("%lld", &s[i]);
        s[n + 1] = INFL;
        n += 2;
        dp[0] = 1, sum[0] = 1;
        int l = 0, r = 0;
        for (int i = 1; i < n; i ++) {
                dp[i] = get(l, r);
                if (i + 1 == n) break;
                while (s[i + 1] - s[r + 1] >= b) r ++;
                if (s[i] - s[i - 1] >= a) ;
                else if (s[i] - s[i - 2] >= a) l = i - 1;
                else l = i;
                (sum[i] = sum[i - 1] + dp[i]) %= MOD;
        }
        long long ans = ((dp[n - 1] % MOD) + MOD) % MOD;
        printf("%lld\n", ans);
        return 0;
}