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>

typedef long long ll;
const int maxn=100000+10,mod=1000000007;
int w[maxn];

ll power(ll a,ll b){
	ll ret=1;
	while(b){
		if(b&1)ret=ret*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ret;
}

ll inv(ll a){
	return power(a,mod-2);
}

int main(){
	// freopen("in.txt","r",stdin);
	int n;ll ans=0;scanf("%d",&n);
	for(int i=1;i<=n;++i)
		scanf("%d",w+i);
	for(int i=1;i<n;++i)
		ans=(ans+(w[n]-w[i])*inv(i))%mod;
	for(int i=1;i<n;++i)
		ans=ans*i%mod;
	printf("%lld\n",ans);
	return 0;
}