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 int long long
const int p=(1e9+7);
int po(int a,int b)
{
    if(b==0)
    {
        return 1;
    }
    if(b==1)
    {
        return a;
    }
    if(b%2==0)
    {
        int u=po(a,b/2);
        return (u*u)%p;
    }
    int u=po(a,b-1);
    return (a*u)%p;
}
int inv(int x)
{
    return po(x,p-2);
}
int32_t main()
{
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int n;
    cin>>n;
    vector <int> a(n);
    for(int i=0;i<n;++i)
    {
        cin>>a[i];
    }
    int ans=0;
    int d[n+2];
    d[0]=0;
    for(int i=1;i<=n;++i)
    {
        d[i]=d[i-1]+inv(i);
        d[i]%=p;
    }
    for(int i=0;i<(n-1);++i)
    {
        ans+=(a[i+1]-a[i])*d[i+1];
        ans%=p;
    }
    for(int i=1;i<=(n-1);++i)
    {
        ans*=i;
        ans%=p;
    }
    cout<<ans;
    return 0;
}
