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<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define p 1000000007
using namespace std;
long long n,m,a[10000001],f[10000001],inv[10000001],cnt,ans;

void line(int x){
	for(int i=2;i<=x;i++)ans*=i,ans%=p;
}

int main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	inv[1]=1;
	for(int i=2;i<n;i++){
		inv[i]=(p-p/i)*inv[p%i]%p;
	}
	for(int i=1;i<n;i++){ 
		f[i]=((1*inv[i]%p*(1+f[i-1]))%p+((i-1)*inv[i]%p*f[i-1])%p)%p;
	}
	for(int i=2;i<=n;i++){
		ans=(ans+(f[i-1]*(a[i]-a[i-1])%p)%p)%p;
	}
	line(n-1);
	cout<<ans%p;
}