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>
typedef long long ll;
using namespace std;
const ll mod = 1000000007;
const ll mod009 = 1000000009;
const ll mod998 = 998244353;
const ll intmax = 2147483647;
const ll llmax = 9223372036854775807;
const char sp = ' ';

int N, x[100000];
ll res;

ll pw(ll x, int y) {
	ll a = 1;
	while (y) {
		if (y & 1)a = a*x%mod;
		x = x*x%mod;
		y /= 2;
	}
	return a;
}

ll modinv(ll x) {
	return pw(x, mod - 2);
}

int main() {
	cin >> N;
	for (int i = 0; i < N; i++) {
		cin >> x[i];
	}
	ll t = 0;
	for (int i = 1; i <= N - 1; i++) {
		t += modinv(i);
		t %= mod;
		res += t*(x[i] - x[i - 1]);
		res %= mod;
	}
	for (int i = 1; i <= N - 1; i++) {
		res *= i;
		res %= mod;
	}
	cout << res << endl;
}