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 MOD = 1000000007;
vector<long long> F;

long long Calc(long long A, long long B) {
	if (B == 0) return 1;
	if (B % 2 == 0) {
		long long C = Calc(A, B / 2);
		return (C * C) % MOD;
	}
	return (A * Calc(A, B - 1)) % MOD;
}

void init(long long N) {
	F.resize(N + 1);
	F[0] = F[1] = 1;
	for (long long i = 2; i <= N; i++) {
		F[i] = (F[i - 1] * i) % MOD;
	}
	return;
}

int main() {
	long long N, ANS = 0, Cur = 0;
	vector<long long> X;
	cin >> N;
	X.resize(N);
	init(N);
	for (int i = 0; i < N; i++) cin >> X[i];
	for (int i = 1; i < N; i++) {
		long long COUNT = (F[N - 1] * Calc(i, MOD - 2)) % MOD;
		Cur = (Cur + COUNT) % MOD;
		ANS = (ANS + Cur * (X[i] - X[i - 1])) % MOD;
	}
	cout << ANS << endl;
}