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 <iostream>

using namespace std;
typedef long long ll;
const int MAX = 100010;
ll inv[MAX],fact[MAX],fi[MAX],mod = 1000000007;

void inverse(){
	int i;
	inv[1] = 1;
	for(i=2;i<MAX;i++){
		inv[i] = mod - (mod/i)*inv[mod%i]%mod;
	}
	fact[0] = fi[0] = 1;
	for(i=1;i<MAX;i++){
		fact[i] = fact[i-1]*i%mod;
		fi[i] = (fi[i-1]*inv[i])%mod;
	}
}

ll comb(int n,int k){
	if(n<0 || k<0 || n<k){
		return 0;
	}else{
		return fact[n]%mod*fi[k]%mod*fi[n-k]%mod;
	}
}

ll x[100010],ans = 0,sum = 0;
int main(){
	int i,n;
	cin >> n;
	for(i=0;i<n;i++){
		cin >> x[i];
	}
	inverse();
	for(i=1;i<n;i++){
		sum += inv[i];
		(ans += sum*(x[i] - x[i - 1])%mod) %= mod;
	}
	(ans *= fact[n - 1]) %= mod;
	cout << ans << endl;
}