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 rep(i,n) for(int i=0;i<n;++i)
typedef long long ll;
ll mod = 1e9+7;
ll modpow(ll a, ll b) {
	ll c = 1;
	while (b > 0) {
		if (b & 1) c = a * c%mod; 
    	a = a * a%mod;
    	b >>= 1;
	}
	return c;
}
int main() {
    ll n;
	cin >> n;
	vector<ll> a(n);
	rep(i, n) { cin >> a[i]; }
	//(N-1)!
	ll fac = 1;
	for(int i=1; i<n; i++) fac = fac * i%mod;
	//mod inverse(逆元)
	//fermatの小定理版
	vector<ll> cnt(n);
	for(int i=1; i<n; i++) {
		cnt[i] = fac * modpow(i, mod - 2) % mod;
		cnt[i] = (cnt[i] + cnt[i - 1]) % mod;
	}
    ll ans = 0;
	rep(i, n-1) {
		ans += (a[i + 1] - a[i]) * cnt[i+1];
		ans %= mod;
	}
	cout << ans << endl;
}
