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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll mod = 1000000007;
#define SIZE 200000
ll inv[SIZE + 1];
ll kai[SIZE + 1];
ll invkai[SIZE + 1];
void invinit()
{
	inv[1] = 1;
	for (int i = 2; i <= SIZE; i++)
	{
		inv[i] = mod - (mod / i)*inv[mod%i] % mod;
	}
	kai[0] = invkai[0] = 1;
	for (int i = 1; i <= SIZE; i++)
	{
		kai[i] = kai[i - 1] * i%mod;
		invkai[i] = invkai[i - 1] * inv[i] % mod;
	}
}
ll com(ll a, ll b)
{
	if (b < 0 || a < b)return 0;
	return (invkai[b] * invkai[a - b]) % mod*kai[a] % mod;
}
int main()
{
	int num;
	scanf("%d", &num);
	invinit();
	ll r = 0;
	int bef;
	ll t = 0;
	for (int i = 0; i < num; i++)
	{
		int z;
		scanf("%d", &z);
		if (i != 0)
		{
			t = (t + kai[num - 1] * inv[i]) % mod;
			r = (r + t*(z - bef)) % mod;
		}
		bef = z;
	}
	printf("%lld\n", r);
}