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 long long p=1e9+7;
long long n;
long long inv[110000];
long long ans=0;
int main(){
	cin>>n;
	inv[1]=1;
	for(long long i=2;i<=n;i++) 
	inv[i]=(p-p/i)*inv[p%i]%p;
	long long f=0;
	long long x1,x2;
	cin>>x1;
	for(long long i=1;i<=n-1;i++){
		cin>>x2;
		f=(f+1)*inv[i]%p+(i-1)*f%p*inv[i]%p;
		ans=(ans+(x2-x1)%p*f%p)%p;
		x1=x2;
	}
	for(int i=1;i<=n-1;i++) ans=(ans*i)%p;
	cout<<ans%p;
}