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 MAXN 100*1000 + 5

int fact[MAXN], inv[MAXN], arr[MAXN];
const int mod = 1000*1000*1000 + 7;

int mul(int a, int b) {
	return (1LL * a * b) % mod;
}

int mpow(int a, int n) {
	if(n == 0) return 1;
	int x = mpow(a, n/2);
	x = mul(x, x);
	if(n % 2) x = mul(x, a);
	return x;
}

void add_self(int& a, int b) { a += b; if(a >= mod) a -= mod; }

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	fact[0] = 1;
	for(int i = 1; i < MAXN; i++) {
		fact[i] = mul(i, fact[i-1]);
		inv[i] = inv[i-1];
		add_self(inv[i], mpow(i, mod-2));
	}
	
	int n;
	cin >> n;
	for(int i = 1; i <= n; i++) {
		cin >> arr[i];
	}
	
	int ans = 0;
	for(int i = 1; i < n; i++) {
		int dist = arr[i+1] - arr[i];
		add_self(ans, mul(dist, inv[i]));
	}
	
	ans = mul(ans, fact[n-1]);
	cout << ans << endl;
}
