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;

#define int long long
#define F first
#define S second

const int MOD = 1000 * 1000 * 1000 + 7;
int32_t main() {
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	int a[n], dis[n], sum[n], fact[n + 10], inv[n + 10];
	fact[0] = fact[1] = 1;
	inv[0] = inv[1] = 1;
	for (int i = 2; i < n + 10; i++) {
		fact[i] = fact[i - 1] * i % MOD;
		inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD;
	}
	for (int i = 0; i < n; i++)
		cin >> a[i];
	int ans = 0;
	sum[0] = 0;
	for (int i = 0; i < n - 1; i++) {
		ans += ((sum[i] + i + 1) * inv[i + 1] % MOD * (a[i + 1] - a[i])) % MOD;
		sum[i + 1] = (sum[i] + (sum[i] + i + 1) * inv[i + 1] % MOD) % MOD;
	}
	ans %= MOD;
	cout << ans * fact[n - 1] % MOD;
}

