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
//by (111110101)2
#include<iostream>
#include<sstream>
#include<fstream>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
#include<functional>
#include<utility>
#include<bitset>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<cstdio>
using namespace std;
const long long md=1e9+7;
long long n,a[101010],qian[101010],hou[101010],ans;
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
	}
	long long bf=1;
	qian[1]=1;
	for(int i=2;i<n;i++)
	{
		bf=bf*(i-1);
		bf%=md;
		qian[i]=(qian[i-1]*i+bf)%md;
	}
	hou[n-1]=1;
	for(int i=n-2;i>=0;i--)
	{
		hou[i]=hou[i+1]*(i+1)%md;
	}
	for(int i=1;i<n;i++)
	{
		ans+=(a[i]-a[i-1])*qian[i]%md*hou[i]%md;
		ans%=md;
	}
	cout<<ans<<endl;
	return 0;
}