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 rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

const ll MOD = 1e9 + 7;
const int MX = 1e5 + 5;

int n;
ll ans;

ll mexp(ll a, ll b) {
	ll res = 1;
	a %= MOD;
	while (b > 0) {
		if (b & 1) {
			res = res * a % MOD;
		}
		a = a * a % MOD;
		b >>= 1;
	}
	return res;
}


int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n;
	vector<ll> a(n);
	for (auto& i : a) cin >> i;
	ll factor = 0;
	rep(i, 1, n) {
		factor += mexp(i, MOD - 2);
		factor %= MOD;
		ans += factor * (a[i] - a[i - 1]);
		ans %= MOD;
	}
	
	rep(i, 1, n) {
		ans = ans * i % MOD;
	}
	cout <<  ans;
}
