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;
const ll MOD = 1000000007;

ll modpow(ll x,ll k){
	if(k == 0)return 1;
	ll ret = modpow(x,k/2);
	ret *= ret; ret %= MOD;
	if(k%2 == 1){
		ret *= x;
		ret %= MOD;
	}
	return ret;
}

int main(){
	static ll n;
	static ll x[100010];
	scanf("%lld",&n);
	for(int i = 0 ; i < n ; i ++){
		scanf("%lld",&x[i]);
	}
	
	ll ret = 0;
	ll sum = MOD-1;
	ll t = modpow(2,MOD-2);
	for(int i = 0 ; i < n-1 ; i ++){
		ret += sum*x[i];
		ret %= MOD;
		if(i > 0){
			t *= i; t %= MOD;
			t *= modpow(i+2,MOD-2); t %= MOD;
		}
		sum += t; if(sum >= MOD)sum -= MOD;
		ll y = modpow(n-1-i,MOD-2);
		y *= x[n-1];
		y %= MOD;
		ret += y;
		if(ret >= MOD)ret -= MOD;
	}
	for(int i = 1 ; i < n ; i ++){
		ret *= i;
		ret %= MOD;
	}
	cout << ret << endl;
}
