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 be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;

int main() {
	cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    ll n;
    cin>>n;
    ll x[n],inv[n];
    for(int i=0;i<n;i++)inv[i]=1;
    for(int i=0;i<n;i++)cin>>x[i];
    for (int i=2;i<n;i++) {
        inv[i] = ll(mod-mod/i)*inv[mod%i]%mod;
    }
    ll ret = 0, sum = 0;
    for (int i = 0; i + 1 < n; ++i) {
        sum+=inv[i + 1];
        sum%=mod;
        ret+=(x[i+1]-x[i])*sum%mod;
    }
    ret%=mod;
    for (int i=1;i<n;i++)ret=ret*i%mod;
    cout << ret<<endl;
    return 0;
}
