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;
 
typedef long long LL;

LL mod = 1000000007;

LL mi (LL x, LL m) {
	// compute pow(x, m-2)
	LL acc = 1;
	LL p = m-2;
	while (p) {
		if (p%2 == 1) {
			acc *= x;
			acc %= m;
		}
		x *= x % m;
		x %= m;
		p >>= 1;
	}
	
	return acc;
}

int main()
{
	LL n;
	cin >> n;

	vector<LL> x(n);

	for (auto &e : x) {
		cin >> e;
	}

	vector<LL> mijs(n-1);

	mijs[0] = 1;
	for (LL i = 1; i < n-1; ++i) {
		mijs[i]=mijs[i-1]+mi(i+1, mod);
		mijs[i]%=mod;
	}

	LL sum = 0;
	for (LL i = 0; i < n-1; ++i) {
		sum += (x[i+1]-x[i])*mijs[i] % mod;
		sum %= mod;
	}

	for (LL i = 1; i <= n-1; ++i) {
		sum *= i;
		sum %= mod;
	}

	cout << sum << "\n";

}