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>
#define ll long long int
#define M 1000000007
#define mod 998244353
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define pi pair<ll,ll>
using namespace std;
const ll N=500010;

ll power(ll x,ll n)
{
    if(n==0){return 1;}
    if(n==1){return x;}
    if(n%2==0){
        return power((x*x)%M,n/2);
    }
    if(n%2==1){
        return (power((x*x)%M,n/2)*x)%M;
    }
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    ll n;
    cin>>n;
    ll ar[n];
    for(ll i=0;i<n;++i)
    {
        cin>>ar[i];
    }

    ll fac[n+1];
    fac[0]=1;
    for(ll i=1;i<=n;++i)
    {
        fac[i]=(i*fac[i-1])%M;
    }

    ll maths[n-1];
    for(ll i=0;i<n-1;++i)
    {
        maths[i]=(fac[n-1]*power(i+1,M-2))%M;
    }
    ll ans=0;
    for(ll i=0;i<n-1;++i)
    {
        ll ch=((ar[n-1]-ar[i])*maths[i])%M;
        ans=(ans+ch)%M;
    }
    cout<<ans;

    return (0);
}
