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
//Author - Rahil Malhotra
#include <bits/stdc++.h>
using namespace std;
 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl '\n'
#define int long long
#define double long double

const int mod=1e9+7;
int inv(int x)
{
    int r,y;
    for(r=1,y=mod-2;y>0;x=(x*x)%mod,y/=2)
        if(y%2==1)
            r=r*x%mod;
    return r;
}
int32_t main()
{
    IOS
    int fact[100005]={};
    fact[0]=1;
    for(int i=1;i<=100002;i++)
        fact[i]=(i*fact[i-1])%mod;
    int n;
    cin>>n;
    int arr[n];
    for(int i=0;i<n;i++)
        cin>>arr[i];
    int st=1;
    int ans=0;
    int sum=0;
    int mul=fact[n-1];
    for(int i=0;i<n-1;i++)
    {
        int a=(mul*inv(i+1))%mod;
        sum+=a;
        sum%=mod;
        ans+=(a*arr[i])%mod;
        ans%=mod;
    }
    ans-=(sum*arr[n-1])%mod;
    if(ans<mod)
        ans+=mod;
    ans=-ans;
    while(ans<0)
        ans+=mod;
    ans%=mod;
    cout<<ans;
}
