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>
#define REG register
using namespace std;
inline void read(int& x){
	static char c;
	while(!isdigit(c=getchar()));x=c^48;
	while(isdigit(c=getchar()))x=(x*10)+(c^48);
}

const int Mod=1000000007;

int n;

int x[100005];

int Pow(int a,int b){
	int ans=1;
	while(b){
    	if(b&1) ans=1ll*ans*a%Mod;
    	a=1ll*a*a%Mod,b>>=1;
    }
	return ans;
}

int _Inv[100005];

inline void Work(){
	read(n);
	for(REG int i=1;i<=n;++i) read(x[i]);
	int Ans=0;
	for(REG int i=1;i<=n;++i) _Inv[i]=(_Inv[i-1]+Pow(i,Mod-2))%Mod;
	for(REG int i=1;i<n;++i) Ans=(Ans+1ll*(x[i+1]-x[i])*_Inv[i]%Mod)%Mod;
	for(REG int i=1;i<n;++i) Ans=1ll*Ans*i%Mod;
	printf("%d\n",Ans);
}

int main(){Work();}