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 rep(i,l,r) for (int i=l,i##end=r;i<=i##end;++i)
#define ll long long
using namespace std;
const int N=1e6+6,P=1e9+7;
int inv[N],x[N];
int main(){
    int n; cin>>n;
    inv[1]=1; rep(i,2,n) inv[i]=(ll)inv[P%i]*(P-P/i)%P;
    rep(i,2,n) inv[i]=(inv[i-1]+inv[i])%P;
    rep(i,1,n) cin>>x[i];
    int ans=0;
    rep(i,1,n-1) ans=(ans+(ll)inv[i]*(x[i+1]-x[i]))%P;
    rep(i,1,n-1) ans=(ll)ans*i%P;
    cout<<ans<<endl;
    return 0;
}
