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 = 1000000007;
long long modpow(long long x, int n) {
	long long r = 1;
	while (n > 0) {
		if (n&1) (r *= x) %= mod;
		n /= 2;
		(x *= x) %= mod;
	}
	return r;
}
long long modinv(long long x) {
	return modpow(x, mod-2);
}
int main() {
	int N;
	cin >> N;
	vector<long long> X(N);
	for (auto& x : X) cin >> x;
	long long aa = 1;
	for (int i = 2; i <= N-1; ++ i) aa = aa * i % mod;
	long long r = 0;
	for (int i = 0; i < N-1; ++ i) {
		r += (X[N-1] - X[i]) * aa % mod * modinv(i+1) % mod;
		r %= mod;
	}
	cout << r << endl;
}
