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>

#define lli long long int
#define ii pair<int, int>
#define ll pair<lli, lli>
#define pb push_back
#define pf push_front
#define pob pop_back
#define pof pop_front
#define mp make_pair
#define fi first
#define se second

using namespace std;

lli bigmod = 1000000007;

lli fex(lli a, lli b) {
	lli ret;
	if (b == 0)
		return 1;
	ret = fex(a, b / 2);
	ret = (ret * ret) % bigmod;
	if (b & 1)
		ret = (ret * a) % bigmod;
	return ret;
}

int main() {
	lli n, x[100005], f, dist, temp, ans = 0, mul = 0;
	scanf("%lld", &n);
	f = 1;
	for (lli i = 2; i < n; i++)
		f = (f * i) % bigmod;
	for (lli i = 0; i < n; i++)
		scanf("%lld", &x[i]);
	for (lli i = 1; i < n; i++) {
		dist = x[i] - x[i - 1];
		temp = (f * fex(i, bigmod - 2)) % bigmod;
		mul = (mul + temp) % bigmod;
		temp = (mul * dist) % bigmod;
		ans = (ans + temp) % bigmod;
	}
	printf("%lld\n", ans);
	return 0;
}