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;
#define F first
#define S second
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;

ll MOD=1e9+7;

ll modinv(ll x)
{
    ll res=1;
    ll pw=MOD-2;
    while(pw>0)
    {
        if(pw%2) res=(res*x)%MOD;
        pw/=2;
        x=(x*x)%MOD;
    }
    return res;
}

ll t[100009];
ll fact=1;
ll n;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    ll sum=0;
    ll ans=0;
    for(ll i=0; i<n; i++)
    {
        cin>>t[i];
        if(i==0) continue;
        fact=(fact*i)%MOD;
        ll va=t[i]-t[i-1];
        sum+=modinv(i);
        ll pl=(sum*va)%MOD;
        ans=(ans+pl)%MOD;
    }
    ans=(ans*fact)%MOD;
    cout<<ans<<endl;
    return 0;
}
