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 int long long
#define pie 3.141592653589793238462643383279
#define mod 1000000007
#define INF LLONG_MAX/5
#define all(v) v.begin(),v.end()
#define P pair<int,int>
#define S second
#define F first
#define f(i,n) for(int i=0;i<n;i++)
int mypow(int a,int b,int c){
	if(b==0)return 1;
	if(b%2)return mypow(a,b-1,c)*a%c;
	int z=mypow(a,b/2,c);
	return z*z%c;
}
int gcd(int a,int b){
	if(b==0)return a;
	return gcd(b,a%b);
}
int lcm(int a,int b){
	return a/gcd(a,b)*b;
}
int x[100000],y[100000],z[100000];
signed main(){
	int n,ans=0;
	cin>>n;
	f(i,n)cin>>x[i];
	z[0]=1;
	f(i,n-1){
		z[i+1]=z[i]*(i+1)%mod;
		y[i+1]=(y[i]*(i+1)+z[i])%mod;
	}
	f(i,n-1){
		ans+=(x[i+1]-x[i])*(z[n-1]*mypow(z[i+1],mod-2,mod)%mod)%mod*y[i+1]%mod;
		ans%=mod;
	}
	cout<<ans<<endl;
}
