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 all(x) (x).begin(),(x).end()
const int mod=1000000007,MAX=200005,INF=1<<30;

ll rui(ll a,ll b){
    ll ans=1;
    while(b>0){
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b/=2;
    }
    return ans;
}

ll fac[MAX];

int main(){
    
    std::ifstream in("text.txt");
    std::cin.rdbuf(in.rdbuf());
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    int N;cin>>N;
    vector<ll> X(N);
    for(int i=0;i<N;i++) cin>>X[i];
    
    fac[0]=1;
    fac[1]=1;
    for(ll i=2;i<MAX;i++){
        fac[i]=fac[i-1]*i;
        fac[i]%=mod;
    }
    
    ll p=1;
    ll sum=0;
    for(int i=1;i<N;i++){
        sum+=(X[i]-X[i-1])*p;
        sum%=mod;
        p+=rui(i+1,mod-2);
        p%=mod;
    }
    
    sum*=fac[N-1];
    sum%=mod;
    
    cout<<sum<<endl;
}

