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 <iostream>
#include <vector>
#include <algorithm>
typedef long long ll;
using namespace std;
#define rep(i, n) for(ll i = 0LL; i < (n); i++)
using Graph = vector<vector<int>>;
ll mod = 1000000007LL;
vector<ll> y;

ll modpow(ll x, int p = mod - 2) {
	if (p == 0) return 1;
	ll temp = modpow(x, p / 2);
	temp = temp * temp % mod;
	if (p % 2) temp = temp * x % mod;
	return temp;
}

int main(void) {
	ll n;
	cin >> n;
	vector<ll> x(n), dif(n), inv(n), inv_sum(n, 0), fact(n, 1);
	rep(i, n) {
		cin >> x[i];
	}
	
	rep(i, n - 1) {
		dif[i + 1] = x[i + 1] - x[i];
	}

	rep(i, n - 1) {
		inv[i + 1] = modpow(i + 1);
		inv_sum[i + 1] = (inv_sum[i] + inv[i + 1]) % mod;		
	}

	rep(i, n - 1) {
		fact[i + 1] = fact[i] * (i + 1) % mod;
	}

	ll ans = 0;
	rep(i, n - 1) {
		ll temp = (inv_sum[i + 1] * dif[i + 1]) % mod;
		ans += temp;
		ans %= mod;
	}
	ans = ans * fact[n - 1] % mod;

	cout << ans << endl;
	return 0;
}