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;
long long fac[100010], inv[100010]; 
long long P = 1000000007;
void init() {
    fac[0] = fac[1] = 1; 
    inv[1] = 1; 
    for (int i = 2; i < 100005; i++) {
        fac[i] = fac[i - 1] * i % P;
        inv[i] = P - inv[P % i] * (P / i) % P;
    }
}
int main(){
	int N; cin >> N; init(); 
	long long X[114514] = {}; 
	for(int i = 0; i < N; i++) cin >> X[i]; 
	long long Y[114514] = {}; 
	for(int i = 0; i < N; i++) Y[i] = X[N-1] - X[i]; 
	long long ans = 0; 
	for(int i = 0; i < N; i++){
		long long tmp = Y[i] * inv[i+1] % P; 
		ans += tmp; ans %= P; 
	}
	ans *= fac[N-1]; ans %= P; 
	cout << ans << endl; 
}