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;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

using ll = long long;

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const int INF = 1<<30;
const ll mod = 1000000007LL;

ll pow(ll a, ll n, ll m) {
	ll ret = 1LL;
	for (; n > 0LL; n >>= 1LL, a = a * a % m) {
		if (n % 2LL == 1LL) {
			ret = ret * a % m;
		}
	}
	return ret;
}

int main() {
    ll N;
    cin>>N;
    vector<ll> v(N);
    for(ll i = 0; i < N; i++) cin>>v[i];
    vector<ll> d(N-1);
    for(ll i = 0; i < N-1; i++) d[i]=v[i+1]-v[i];
    ll ans = 0LL;
    ll P = 1;
    for(ll i = 1;i < N; i++) P = (P*i)%mod;
  	ll Q = P;
    for(ll i = 0; i < N-1; i++){
        ans=(ans+(d[i]*P)%mod)%mod;
        P=(P+Q*pow(i+2LL,mod-2,mod))%mod;
    }
    cout<<ans<<endl;
}