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 MOD=(int)1e9+7;
const int MAX=510000;

long long fac[MAX],finv[MAX],inv[MAX];

//テーブルを作る前処理
void COMinit(){
  fac[0]=fac[1]=1;
  finv[0]=finv[1]=1;
  inv[1]=1;
  for(int i=2;i<MAX;i++){
    fac[i]=fac[i-1]*i%MOD;
    inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
    finv[i]=finv[i-1]*inv[i]%MOD;
  }
}

//二項係数の計算
long long COM(int n,int k){
  if(n<k) return 0;
  if(n<0 || k<0)  return 0;
  return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;
}

int main(){
	int N;
	cin >> N;
	vector<int> x(N);
	for(int i=0;i<N;i++)	cin >> x.at(i);

	COMinit();
	long long X=1;//(N-1)!
	for(int i=1;i<N;i++)	X=(X*i)%MOD;
	vector<long long> C(N,0);
	C.at(0)=0;
	for(int i=1;i<N;i++){
		C.at(i)=C.at(i-1) + X*inv[i];
		C.at(i)%=MOD;
	}
	long long ans=0;
	for(int i=1;i<N;i++){
		ans+=C.at(i)*(x.at(i)-x.at(i-1));
		ans%=MOD;
	}
	cout << ans << endl;

	return 0;
}
