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 <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <functional>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cassert>

#define SIZE 1000000007
 
using namespace std;
typedef long long int ll;
typedef pair <int,int> P;
 
using namespace std;


int main()
{	
	ll N;
	scanf("%lld", &N);
	ll x[N];
	
	for(ll i=0;i<N;i++){
		scanf("%lld", &x[i]);
	}
	ll kai[N];
	kai[0]=1;
	for(ll i=1;i<N;i++){
		kai[i]=kai[i-1]*i;
		kai[i]=kai[i] % SIZE;
	}
	
	ll answer=0;
	ll np=1;
	for(ll i=1;i<N;i++){
		answer=answer+kai[i-1]*(x[N-1]-x[i-1]);
		answer=answer % SIZE;
		if(i!=N-1)
		{answer=answer*(i+1);
		 answer=answer % SIZE;
		}
	
	}
	
	printf("%lld", answer);
}
