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 int long long
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB push_back
#define PF push_front
#define FILL(a,b) memset(a,b,sizeof(a))
using namespace std;
int f1[100005],f2[100005],a1[100005],a2[100005];
int n,a[100005],mod=1e9+7;
signed main(){
	ios::sync_with_stdio(0);
    cin.tie(0);
	int ans=0;
	cin>>n;
	FOR(i,1,n){
		cin>>a[i];
	}
	FOR(i,1,n){
		a1[i]=(a1[i-1]+a[i])%mod;
	} 
	for(int i=n-1;i>=1;i--){
		a2[i]=(a2[i+1]+a[i])%mod;
	}
	f1[0]=1;
	FOR(i,1,n-1){
		f1[i]=f1[i-1]*i%mod;
	} 
	f2[n-1]=n-1;
	for(int i=n-2;i>=1;i--){
		f2[i]=f2[i+1]*i%mod;
	}
	f2[n]=f2[n+1]=1;
	for(int i=1;i<n-1;i++){
		ans=ans+f2[i+2]*f1[i-1]%mod*(a2[i+1]-a1[n-i-1]);
		ans=ans%mod;
	}
	FOR(i,1,n-1){
		ans=ans+f1[n-i-1]*f2[n-i+1]%mod*(a[n]-a[i]);ans=ans%mod;
	}
	cout<<ans;
	return 0;
}


