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 ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

int const N = 100000, md = 1e9 + 7;
int fc[N + 1], inv[N + 1], x[N + 1];

void ad(int &x, int y) { if ((x += y) >= md)x -= md; }
void sb(int &x, int y) { if ((x -= y) <  0 )x += md; }
int sm(int l, int r){
	int an = x[r];
	sb(an, x[l - 1]);
	return an;
}

int main(){
	int n;
	scanf("%d", &n);
	
	fc[0] = 1;
	f(i, 1, n + 1)fc[i] = (ll)fc[i - 1] * i % md;
	inv[1] = 1;
	f(i, 2, n + 1)inv[i] = md - md / i * (ll)inv[md % i] % md;
	
	f(i, 1, n + 1)scanf("%d", x + i);
	
	int an = 0;
	f(i, 1, n)ad(an, (ll)(x[n] - x[i]) * fc[n - 1] % md * inv[n - i] % md);
	
	f(i, 1, n + 1)ad(x[i], x[i - 1]);
	
	--n;
	f(i, 0, n - 1){
		int s = sm(2 + i, n);
		sb(s, sm(1, n - 1 - i));
		ad(an, (ll)s * fc[n] % md * inv[i + 1] % md * inv[i + 2] % md);
	}
	
	printf("%d\n", an);
}