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;

#define MOD (1000000000+7)

long long power( long long x, long long n )
{
	long long res = 1;
	while( n > 0 ) {
		if( n % 2 ) res = res * x % MOD;
		x = x * x % MOD;
		n /= 2;
	}
	return res;
}

int main()
{
	int N;
	cin >> N;
	vector<long long> x( N );
	for( int i = 0; i < N; i++ ) {
		cin >> x[i];
	}

	long long fac = 1;
	for( int i = 2; i < N; i++ ) {
		fac *= i;
		fac %= MOD;
	}
	vector<long long> p( N );
	p[0] = fac;
	for( int i = 1; i < N; i++ ) {
		p[i] = p[i - 1] + fac * power( i + 1, MOD - 2 ) % MOD;
		p[i] %= MOD;
	}
	long long ans = 0;
	for( int i = 0; i < N - 1; i++ ) {
		ans += (x[i + 1] - x[i]) * p[i] % MOD;
		ans %= MOD;
	}

	cout << ans << endl;
}
