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;
using ll = long long;

const int MOD = 1e9+7;
const int MAX_N = 1e5+12;
int n;
ll x[MAX_N];
ll prob[MAX_N];

ll powmod(ll x, ll y)
{
	ll res = 1;
	while(y){
		if(y&1) (res*=x)%=MOD;
		(x*=x)%=MOD;
		y>>=1;
	}
	return res;
}

void solve()
{
	prob[1]=1;
	for(int i=2;i<n;++i) prob[i]=(prob[i-1]+powmod(i,MOD-2))%MOD;
	ll ans = 0;
	for(int i=1;i<n;++i) (ans+=(x[i]-x[i-1])*prob[i])%=MOD;
	for(int i=1;i<n;++i) (ans*=i)%=MOD;
	cout<<ans<<'\n';
}

int main()
{
	cin >> n;
	for(int i=0;i<n;++i) cin >> x[i];
	solve();
	return 0;
}