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;

#define MOD 1000000007

int n;
long long fa[100005], fb[100005], fc[100005], la, ans;

int main() {
	scanf("%d", &n);
	fa[0] = 1;
	for (int i = 1; i < n; i++)
		fa[i] = fa[i - 1] * i % MOD;
	fb[n - 1] = 1;
	for (int i = n - 1; i; i--)
		fb[i - 1] = fb[i] * i % MOD;
	for (int i = 1; i < n; i++)
		fc[i] = (fc[i - 1] + fa[i - 1] * fb[i]) % MOD;
	scanf("%lld", &la);
	for (int i = 1; i < n; i++) {
		long long tmp;
		scanf("%lld", &tmp);
		// printf("%lld %lld\n", tmp - la, fc[i]);
		ans = (ans + (tmp - la) * fc[i]) % MOD;
		la = tmp;
	}
	printf("%lld\n", ans);
}