There are N slimes standing on a number line. The i-th slime from the left is at position x_i.

It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.

Niwango will perform N-1 operations. The i-th operation consists of the following procedures:

* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.



Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.

Constraints

* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.

Input

Input is given from Standard Input in the following format:


N
x_1 x_2 \ldots x_N


Output

Print the answer.

Examples

Input

3
1 2 3


Output

5


Input

12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932


Output

750927044
#include <iostream>
using namespace std;

const int M = 1000000007;
long long modinv(int a, int m) {
    int x = m, y = a, p = 1, q = 0, r = 0, s = 1;
    while (y != 0) {
        int u = x / y;
        int x0 = y; y = x - y * u; x = x0;
        int r0 = p - r * u, s0 = q - s * u;
        p = r; r = r0; q = s; s = s0;
    }
    return q < 0 ? q + m : q;
}

int main() {
    int n; cin >> n;
    int x[n]; for (int i = 0; i < n; i++) cin >> x[i];
    long long f = 1, A = 0, ans = 0;
    for (int i = 1; i < n; i++) f = f * i % M;
    for (int i = 1; i < n; i++) {
        A += f * modinv(i, M) % M;
        ans += A * (x[i] - x[i-1]) % M;
    }
    cout << ans % M << endl;
}
