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 <bits/stdc++.h>
using namespace std;

const int N = 100005, P = 1000000007;
int n, x[N], h[N];

int power(int a, int x) {
	int ans = 1;
	for (; x; x >>= 1, a = 1ll * a * a % P)
		if (x & 1) ans = 1ll * ans * a % P;
	return ans;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) scanf("%d", x + i), h[i] = (h[i - 1] + power(i, P - 2)) % P;
	int ans = 0;
	for (int i = 1; i < n; ++i) ans = (ans + 1ll * (x[i + 1] - x[i]) * h[i]) % P;
	for (int i = 1; i < n; ++i) ans = 1ll * ans * i % P;
	printf("%d\n", ans);
}
