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;

long long u[100010], v[100010], w[100010];

int main()
{
	int n, a[100010];
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}

	u[0] = 1;
	int mod = 1000000007;
	for (int i = 1; i <= n; i++) {
		u[i] = u[i - 1] * i;
		u[i] %= mod;
	}

	v[0] = 1;
	for (int i = 1; i < n - 1; i++) {
		v[i] = v[i - 1] * (i + 1) + u[i];
		v[i] %= mod;
	}

	w[n - 1] = 1;
	for (int i = n - 2; i >= 0; i--) {
		w[i] = w[i + 1] * (i + 1);
		w[i] %= mod;
	}

	long long ans = 0;
	long long t;
	for (int i = 0; i < n - 1; i++) {
		t = (a[i + 1] - a[i]) * w[i + 1];
		t %= mod;
		t *= v[i];
		t %= mod;
		ans += t;
		ans %= mod;
	} 
	cout << ans << endl;

	return 0;
}