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<ll,ll> pi;
typedef complex <double> cp;
#define debug(a) cout<<#a<<":"<<a<<endl;
#define fr freopen("in.txt","r",stdin);
#define Fill(x,a) memset(x,a,sizeof(x))
#define cpy(a,b) memcpy(a,b,sizeof(a))
const double PI = acos(-1);
const ll INF=0x3f3f3f3f;
const ll N=1e6+7;
const ll mod=1e9+7;
ll maxn,minn;
ll T,n,m,q;
ll inv[N]; 
ll arr[N];

int main(){
	cin >> n;
	for(ll i = 1; i <= n; i ++ ){
		scanf("%lld", arr + i);
	}
	inv[1] = 1;
	for(ll i = 2; i <= n; i ++ ){
		inv[i] = (mod - mod / i) * inv[mod % i] % mod;
	}
	for(ll i = 2; i <= n; i ++ ){
		inv[i] = (inv[i] + inv[i - 1]) % mod;
	}
	ll ans = 0;
	for(ll i = 1; i < n; i ++ ){
		ans = ans + (arr[i + 1] - arr[i]) * inv[i] % mod;
	}
	for(ll i = 1; i < n ; i ++ ){
		ans = (ans * i ) % mod;
	}
	cout << ans << endl;
	
		
	
	


	return 0;
}

