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;
const int max_n=100005;
const int mod=1e9+7;
inline int Add(int x,int y){
    return (x+=y)>=mod?x-mod:x;
}
inline int Sub(int x,int y){
    return (x-=y)>=0?x:x+mod;
}
inline int Mul(int x,int y){
    return 1ll*x*y%mod;
}
int n,x[max_n];
int inv[max_n];
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",x+i);
    inv[1]=1;
    for(int i=2;i<=n;i++)inv[i]=Mul(inv[mod%i],mod-mod/i);
    for(int i=1;i<=n;i++)inv[i]=Add(inv[i-1],inv[i]);
    int ans=0;
    for(int i=1;i<n;i++){
        int tmp=Sub(x[i+1],x[i]);
        tmp=Mul(tmp,inv[i]);
        ans=Add(ans,tmp);
    }
    for(int i=1;i<n;i++)ans=Mul(ans,i);
    printf("%d\n",ans);
    return 0;
}