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;
const int MAXN=100010,P=1e9+7;
int n,ans,x[MAXN],inv[MAXN];
int qpow (int a,int b) {
	int res=1;
	while (b) {
		if (b&1) {res=(1ll*res*a)%P;}
		a=(1ll*a*a)%P,b>>=1;
	}
	return res;
}
int main () {
	scanf("%d",&n);
	for (int i=1;i<=n;i++) {scanf("%d",&x[i]);}
	for (int i=1;i<=n;i++) {inv[i]=qpow(i,P-2);}
	for (int i=1;i<=n-1;i++) {ans=(ans+(1ll*(x[n]-x[i])*inv[n-i])%P)%P;}
	int tmp=0;
	for (int i=2;i<=n-1;i++) {
		tmp=(tmp+(1ll*inv[i]*inv[i-1])%P)%P;
		ans=(ans+(1ll*x[i]*tmp)%P)%P;
	}
	tmp=0;
	for (int i=n-2;i>=1;i--) {
		tmp=(tmp+(1ll*inv[n-1-i]*inv[n-1-i+1])%P)%P;
		ans=(ans-(1ll*x[i]*tmp)%P+P)%P;
	}
	for (int i=1;i<=n-1;i++) {ans=(1ll*ans*i)%P;}
	printf("%d\n",ans);
	return 0;
}