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>
using namespace std;
int a[100010];
const int P=1000000007; 
int pow(int a,int k){
	if(k==1) return a;
	long long t=pow(a,k/2);
	t%=P;
	t*=t;
	t%=P;
	if(k%2) t*=a;
	return t%P;
}
int main(){
	int n;
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i];
	long long ans=0,sum=0;
	for(int i=1;i<n;i++){
		sum+=pow(i,P-2);
		sum%=P;
		ans+=1LL*sum*(a[i+1]-a[i])%P;
		ans%=P;
	}
	for(int i=1;i<n;i++){
		ans*=i;
		ans%=P;
	}
	cout<<ans;
	return 0;
}