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>
#include <vector>

using namespace std;

const int MOD = 1000000007;

int main(){
	vector<long long> inv(100000);
	vector<long long> fact(100000);
	for(int i=0;i<2;i++) inv[i] = fact[i] = 1;
	for(int i=2;i<inv.size();i++){
		inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD;
		fact[i] = fact[i-1] * i % MOD;
	}
    int N;
    while(cin >> N){
        vector<int> x(N);
        for(auto& t : x) cin >> t;
        long long res = 0;
        long long c = 0;
        for(int i=1;i<N;i++){
            c += fact[N-1] * inv[i];
            c %= MOD;
            res += c * (x[i] - x[i-1]);
            res %= MOD;
        }
        cout << res << endl;
    }
}