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;
typedef pair<int,int> P;

ll modpow(ll a, ll p, ll mod){
        if(p == 0) return 1;
        if(p%2 == 0){
            ll half = modpow(a, p/2,mod);
            return half * half % mod;
        }
        else return a * modpow(a,p-1,mod) % mod;
    }

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll N; cin >> N;
    ll x[100005];
    for (int i = 0;i < N;i++){
        cin >> x[i];
    }

    ll mod = 1e9+7;
    ll ans = 0;

    ll tmp = 1;
    for (ll i = 1;i < N;i++){
        tmp = (tmp*i)%mod;
    }

    ll dif = tmp*modpow(2,mod-2,mod)%mod;

    ll cnt = 2;
    for (ll i = 0;i < N-1;i++){
        //cout << tmp << endl;
        ans += (x[i+1]-x[i])*tmp;
        ans %= mod;
        tmp = (tmp+dif) % mod;
        dif = ((dif*cnt)%mod)*modpow(cnt+1,mod-2, mod)%mod; 
        cnt++;
    }

    cout << ans << endl;

    return 0;

}