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;
#define lef(x) (x<<1)
#define rig(x) (lef(x)+1)
#define int long long
const ll MOD=1e9+7LL;
const int N=100100;
ll md(ll x){
	return ((x%MOD)+MOD)%MOD;
}
ll fastexpo(ll x, ll n)
{
	ll b=1LL;
	while(n)
	{
		if(n&1LL)
			b=md(b*x);
		x=md(x*x);
		n>>=1LL;
	}
	return b;
}
ll invfat[N],fat[N],a[N];
int32_t main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    fat[0]=1;
    invfat[0]=1;
    for (int i = 1; i < N; ++i)
    {
    	fat[i]=md(fat[i-1]*i);
    	invfat[i]=fastexpo(fat[i],MOD-2);
    }
    int n;
    cin>>n;
    for (int i = 1; i <= n; ++i)
    {
    	cin>>a[i];
    }
    ll tot=0LL;
    ll rs=0LL;
    for (int i = 1; i <n; ++i)
    {
    	tot=md(tot+fastexpo(i,MOD-2));
    	// cout<<tot<<" "<<(a[i+1]-a[i])<<"\n";
    	rs=md(rs+(a[i+1]-a[i])*tot);
    }
    // cout<<rs<<"\n";
    rs=md(rs*fat[n-1]);
    cout<<rs<<"\n";
    return 0;   
}