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,n) for(int i=0; i<(n); i++)
using ll = long long;
constexpr int MOD = 1e9+7;
int n, x, p;
ll ac, m, ans;
ll inv(ll n){
	ll res = 1;
	for(ll m = MOD-2; m; m/=2,(n*=n)%=MOD) if(m%2) (res *= n) %= MOD;
	return res;
}
int main(){
	scanf("%d%d", &n, &p);
	rep(i,n-1){
		scanf("%d", &x);
		p = x - p;
		(m += inv((ll)i * (i+1) % MOD)) %= MOD;
		(ac += 1 - m + MOD) %= MOD;
		(ans += ac * p % MOD) %= MOD;
		p = x;
	}
	rep(i,n-1) (ans *= i+1) %= MOD;
	printf("%lld\n", ans);
}