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>
#include <vector>

using namespace std;

int main() {
    int n;
    cin >> n;

    vector<long long> x(n);
    for (int i = 0; i < n; i++) {
        cin >> x[i];
    }

    long long mod = 1000000007;

    vector<long long> inv(n + 1, 0), inv_sum(n + 1, 0), frac(n + 1, 0);
    inv[1] = 1;
    inv_sum[1] = 1;
    frac[0] = frac[1] = 1;
    for (int i = 2; i <= n; i++) {
        inv[i] = mod - inv[mod % i] * (mod / i) % mod;
        inv_sum[i] = (inv_sum[i - 1] + inv[i]) % mod;
        frac[i] = frac[i - 1] * i % mod;
    }

    long long res = 0;
    for (int i = 0; i < n - 1; i++) {
        res += (x[i + 1] - x[i]) * inv_sum[i + 1] % mod * frac[n - 1];
        res %= mod;
    }

    cout << res << endl;
}
