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;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000000

//aのb乗
int beki(int a,int b){
	int x = 1;
	while(b!=0){
		if(b&1){
			x=mod(x*a);
		}
		a=mod(a*a);
		b>>=1;
	}
	return x;
}

//aの逆元
int gyakugen(int a){
	return beki(a,modulo-2);
}

int main(){
	
	int N;
	cin>>N;
	
	vector<int> x(N);
	
	for(int i=0;i<N;i++){
		cin>>x[i];
	}
	
	vector<int> a(N+1,0);
	int sum = 0;
	for(int i=1;i<=N;i++){
		a[i] = mod(sum + i);
		a[i] = mod(a[i] * gyakugen(i));
		sum = mod(sum + a[i]);
	}
	
	int ans = 0;
	
	int S = 0;
	/*
	k = 1;
	for(int i=1;i<=N-1;i++){
		k = mod(k*i);
	}*/
	
	for(int i=0;i<N-1;i++){
		int sum = (a[i+1]);
		sum = mod(sum * (x[i+1]-x[i]));
		ans = mod(ans + sum);
	}
	
	for(int i=1;i<=N-1;i++){
		ans = mod(ans * i);
	}
	
	cout<<ans<<endl;
	
    return 0;
}
