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;

typedef long long ll;

// (a^b)%m
ll exp(ll a, ll b, ll m) {
	ll r=1;
	while(b) {
		if(b%2) r=(r*a)%m;
		a=(a*a)%m;
		b/=2;
	}
	return r;
}

// a^-1 %m, prime m
ll invp(ll a, ll m) {
	return exp(a, m-2, m);
}

const int mx=1e5+77;
const int mod=1e9+7;

ll fact[mx];

int main() {
	fact[0]=1;
	for (int i=1; i<mx; i++)
		fact[i]=(fact[i-1]*i)%mod;
	ll inv[mx];
	int N; cin>>N;
	for (int i=1; i<=N-1; i++)
		inv[i]=(fact[N-1]*invp(i,mod))%mod;
	vector<int> x(N+1);
	for (int i=1; i<=N; i++)
		cin>>x[i];
	ll ret=0;
	for (int i=1; i<=N-1; i++)
		ret=(ret+inv[i]*(x[N]-x[i]))%mod;
	cout<<ret<<"\n";
	return 0;
}