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 M = 1e9+7;

long long pwr( long long a , long long b , long long m = M ){
	
	long long ret = 1;
	a %= m , b %= m;

	while( b>0 ){
		if(b&1)	ret = ret*a % m;
		a = a*a % m;
		b >>= 1;
	}

	return ret;
}

int main(){

	ios_base::sync_with_stdio(0);
	cout.tie(0);
	cin.tie(0);

	int n;
	cin>>n;

	long long fac=1;	
	vector<int> pre(n+5,0);
	for(int i=1 ; i<n ; ++i){
		pre[i] = ( pre[i-1] + pwr(i,M-2) ) % M;
		fac = fac * i % M;
	}

	long long ans=0;
	vector<int> a(n);

	for(int i=0 ; i<n ; ++i){
		cin>>a[i];
		if(i)	ans = ( ans + 1l*(a[i]-a[i-1])*pre[i] % M ) % M;
	}

	ans = ans * fac % M;
	cout<< ans ;
}

