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 = 1e9 + 7;
int N , arr[100003];

int poww(long long a , int b){
	int times = 1;
	while(b){
		if(b & 1) times = times * a % MOD;
		a = a * a % MOD; b >>= 1;
	}
	return times;
}

int main(){
	cin >> N;
	for(int i = 1 ; i <= N ; ++i) cin >> arr[i];
	int sum = 1 , ans = 0;
	for(int i = 2 ; i <= N ; ++i){
		ans = (ans + 1ll * sum * (arr[i] - arr[i - 1])) % MOD;
		sum = (sum + poww(i , MOD - 2)) % MOD;
	}
	for(int i = 1 ; i < N ; ++i) ans = 1ll * ans * i % MOD;
	cout << ans;
	return 0;
}