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 rep(i,n) for(ll i=0;i<n;i++)
#define all(v) v.begin(), v.end()
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<ll> vec;
typedef vector<vec> mat;
ll mod=1e9+7;
ll inv(ll a){
    ll n=mod-2,ans=1;
    while(n>0){
        if(n&1) ans=ans*a%mod;
        n=n/2;
        a=a*a%mod;
    }
    return ans;
}
int main(){
    ll n,x[200001],ans=0,a=0,b=1,mod=1e9+7;
    cin>>n;
    rep(i,n) cin>>x[i];
    rep(i,n-1) (b*=i+1)%=mod;
    rep(i,n-1) a+=b*inv(i+1)%mod,(ans+=(x[i+1]-x[i])*a)%=mod;
    cout<<ans<<"\n";
}