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;
using ll=long long;
const int MOD=(int)1e9+7;
const int INF=(int)1e9;
const ll LINF=(ll)1e18;

ll modpow(ll x,ll n){
    ll res=1;
    while(n>0){
        if(n&1)res=res*x%MOD;
        x=x*x%MOD;
        n>>=1;
    }
    return res;
}

int main(){
    int n;
    cin>>n;
    ll x[n];
    for(int i=0;i<n;i++){
        cin>>x[i];
    }
    ll ans=0;
    ll sum=0;
    for(int i=0;i<n-1;i++){
        sum+=modpow(i+1,MOD-2);
        sum%=MOD;
        ans+=sum*(x[i+1]-x[i]);
        ans%=MOD;
    }
    for(int i=1;i<n;i++){
        ans=(ans*i)%MOD;
    }
    cout<<ans<<endl;
}
