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 debug(x) cerr << #x << " " << (x) << endl
using namespace std;

const int N = 100005;
const long long mod = 1000000007ll;

int n;
long long x[N], inv[N], fac[N], ans = 0ll;

int main () {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%lld", &x[i]);

	fac[0] = 1ll;
	for (int i = 1; i < n; i++) fac[i] = fac[i - 1] * i % mod;

	inv[1] = 1ll;
	for (int i = 2; i <= n; i++) inv[i] = (mod / i) * (mod - inv[mod % i]) % mod;

	long long now = 0ll;
	for (int i = 1; i < n; i++) now = (now + inv[i]) % mod, ans = (ans + now * fac[n - 1] % mod * (x[i + 1] - x[i]) % mod) % mod;
	printf("%lld\n", ans);
	return 0;
}