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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<unordered_set>
#include<unordered_map>
using namespace std;
typedef long long ll;
#define chmax(a,b) a=max(a,b)
#define chmin(a,b) a=min(a,b)
#define all(x) x.begin(),x.end()
#define rep(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
#define mad(a,b) a=(a+b)%mod
#define N 200010
ll po(ll x,ll y){
    ll res=1;
    for(;y;y>>=1){
	if(y&1)res=res*x%mod;
	x=x*x%mod;
    }
    return res;
}
ll inv(ll x){
    return po(x,mod-2);
}
ll fac[N];
ll n,x[N];
int main(){
    cin.tie(0);
    ios::sync_with_stdio(0);
    fac[0]=1;
    for(ll i=1;i<N;i++)fac[i]=fac[i-1]*i%mod;
    cin>>n;
    rep(i,n)cin>>x[i];
    ll ans=0,rui=0;
    for(ll i=1;i<n;i++){
	ll d=x[i]-x[i-1];
	mad(rui,inv(i)%mod*fac[n-1]%mod);
	mad(ans,d*rui);
    }
    cout<<ans<<endl;
}
