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> 
#define ll long long
using namespace std;
const ll p = 1e9 + 7, N = 1e5 + 5;
ll x[N], f[N], ans = 0;
int n;
ll solve (ll x) {
	ll ans = 1;
	ll b = p - 2;
	while (b) {
		if (b & 1) ans = ans * x % p;
		x = x * x % p;
		b >>= 1;
	}
	return ans;
}
int main () {
	scanf ("%d", &n);
	for (int i=1;i<=n;i++) scanf ("%lld", &x[i]);
	f[0] = 0;
	for (int i=1;i<n;i++) {
		ll i_1 = solve (1LL * i);
		f[i] = (i_1 % p * (1 + f[i - 1])
		+ (i - 1) * i_1 % p * f[i - 1]) % p;
//		cout << f[i] << endl;
	}
	for (int i=2;i<=n;i++) {	
		ans = (ans + f[i - 1] * (x[i] - x[i - 1]) % p);
	}
	for (int i=1;i<n;i++) {
		ans = (ans * i) % p;
	}
	cout << ans << endl;
}