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;

const long long MOD = 1000000007;

long long mod_pow(long long x, long long n, long long mod){
	long long res = 1;
	while(n > 0){
		if(n & 1) res = res * x % mod;
		x = x * x % mod;
		n >>= 1;
	}
	return res;
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	long long n; cin >> n;
	long long x[n];
	for(int i = 0; i < n; i++){
		cin >> x[i];
	}
	long long ans = 0, factor = 0;
	for(int i = 1; i < n; i++){
		factor += mod_pow(i, MOD-2, MOD);
		factor %= MOD;
		ans += factor * (x[i] - x[i-1]);
		ans %= MOD;
	}
	for(int i = 1; i <= n-1; i++){
		ans = ans * i % MOD;
	}
	cout << ans << endl;
	return 0;
}