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>
#include <array>
using namespace std;
using ULL = unsigned long long;
using UL = unsigned;
using LL = long long;
#define rep(i, n) for(UL i = 0; i < (n); i++)

struct Problem {

	static const ULL M = 1000000007;
	ULL X[100000];
	ULL fac[100001];
	ULL inv[100001];
	ULL dp[100000];

	void Solve() {
		ULL N; cin >> N;
		rep(i, N) cin >> X[i];
		fac[0] = 1;
		for (ULL i = 1; i <= N; i++) fac[i] = fac[i - 1] * i % M;
		inv[1] = 1;
		for (ULL i = 2; i <= N; i++) inv[i] = (M - inv[M % i] * (M / i) % M) % M;
		ULL ans = 0;
		ULL G = 0;
		for (ULL i = 1; i < N; i++) {
			G = (G + inv[i]) % M;
			ans += (X[i] - X[i - 1]) * G;
			ans %= M;
		}
		ans *= fac[N - 1];
		ans %= M;
		cout << ans << endl;
	}

	Problem();
};
int main() {
	unique_ptr<Problem> p(new Problem());
	p->Solve();
	return 0;
}
Problem::Problem() {
	cout << fixed << setprecision(10);
}
