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 For(i,n,k) for(ll i=(n);i<(k);i++)
#define ALL(a)  (a).begin(),(a).end()
ll ans=0;
const ll MAX = 510000;
const ll MOD = 1000000007;

ll fac[MAX], finv[MAX], inv[MAX];

void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
void Main(){
    ll n;cin>>n;
    vector<ll> x(n);
    For(i,0,n) cin>>x[i];
    COMinit();
    vector<ll> acc(n+1,0);
    For(i,0,n) acc[i+1]=acc[i]+inv[i+1];
    For(i,0,n-1){
        ans+=(x[i+1]-x[i])*acc[i+1]%MOD;
        ans%=MOD;
    }
    ans*=fac[n-1];
    cout<<ans%MOD<<endl;
}
int main(){
    Main();
    return 0;
}