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 <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
int n;
ll fac[100010];
ll x[100010];
ll fans = 0;

ll kmi (ll x, ll t) {
	ll ans = 1;
	for (x %= mod; t; t >>= 1, x = (x * x) % mod)
		if (t % 2)
			(ans *= x) %= mod;
	return ans;
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n;
	fac[0] = 1;
	for (int i = 1; i <= n; i ++)
		fac[i] = fac[i - 1] * i % mod;
	ll cnt = 0;
	for (int i = 1; i <= n; i ++) {
		cin >> x[i];
		if (i == 1)
			continue ;
		(cnt += fac[n - 1] * kmi (i - 1, mod - 2) % mod) %= mod;
		(fans += cnt * (x[i] - x[i - 1]) % mod) %= mod;
	}
	cout << fans << endl;
	return 0;
}