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>

#define ff first
#define ss second
#define pb push_back

using namespace std;
using ll = long long;
using ii = pair<int, int>;

const int N = 100005;
const int mod = 1e9+7;

int n;
ll fat[N];

ll inv(ll b){
	ll ans = 1;
	ll e = mod - 2;
	while(e){
		if(e&1){
			ans = (ans * b) % mod;
		}
		e >>= 1;
		b = (b * b) % mod;
	}
	return ans;
}

int main() {
	fat[0] = 1;
	for(int i = 1; i < N; i++){
		fat[i] = (fat[i-1] * i) % mod;
	}
	ll v[N];
	scanf("%d", &n);
	for(int i = 0; i < n; i++){
		scanf("%lld", v+i);
	}
	ll ans = 0;
	n--;
	ll ant = 0;
	for(int i = 0; i < n; i++){
		ll val = v[i+1] - v[i];
		ant = ant + (fat[n] * inv(i+1)) % mod;
		ant %= mod;
		ans += (val * ant) % mod;
		ans %= mod;
	}
	printf("%lld\n", ans);
	return 0;
}
