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 <stdio.h>
long long p = 1000000007;
long long n, ans;
long long x[100001];
long long y[100001];
long long rui(long long rui_a, long long rui_b) {
	if (rui_b == 0)return 1;
	if (rui_b == 1)return rui_a % p;
	long long rui_c;
	rui_c = rui(rui_a, rui_b / 2);
	rui_c = rui_c * rui_c % p;
	if (rui_b % 2) {
		return rui_c * rui_a % p;
	}
	return rui_c;
}
int main() {
	scanf("%lld", &n);
	for (long long i = 1; i <= n; i++) {
		scanf("%lld", x + i);
	}
	y[1] = 1;
	for (long long i = 1; i < n; i++) {
		y[1] = (i * y[1]) % p;
	}
	for (long long i = 2; i < n; i++) {
		y[i] = y[1] * rui(i, p - 2) % p;
	}
	for (long long i = 1; i < n; i++) {
		ans += (((x[n] - x[i]) * y[i]) % p);
		ans = ans % p;
	}
	printf("%lld\n", ans % p);
}