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 lli long long int
#define REP(i,s,n) for(int i=s;i<n;i++)
#define MOD 1000000007
#define NUM 2520
#define INF (1LL<<50)
#define DEBUG 1
#define mp(a,b) make_pair(a,b)

lli inv[200200];

lli mod_pow(lli a,lli b){
	a %= MOD;
	if(b==0)return 1;
	if(b==1)return a;
	lli res = mod_pow(a,b/2)%MOD;
	res *= res;
	res %= MOD;
	if(b%2)res *=a;
	return res%MOD;
}

signed main(){

	lli n;
	cin>>n;
	vector<lli> x(n);
	REP(i,0,n)cin>>x[i];

	//invMake;
	inv[0]=1;
	REP(i,1,150150)inv[i] = inv[i-1]*i%MOD;

	vector<lli> cnt(n+5,0);
	REP(i,1,n+4){
		cnt[i] = inv[n-1] * mod_pow(i,MOD-2)%MOD;
		cnt[i] += cnt[i-1];
		cnt[i] %= MOD;
	}

	lli ans = 0;
	REP(i,0,n-1){
		ans += (x[i+1]-x[i])*cnt[i+1]%MOD;
		ans %=MOD;
	}
	cout<<ans<<endl;

	return 0;
}
