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"
#pragma warning(disable:4996)
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
const int mod = 1000000007;
const long long inf = 1ll << 61;
typedef pair<int, int> P;
typedef pair<int, P>PP;
int x[100006];
int kj[100006];
int c[100006];

int modpow(int a, int x, int mod) {
	int res = 1;
	while (x) {
		if (x & 1)res = res*a%mod;
		a = a*a%mod;
		x >>= 1;
	}
	return res;
}

signed main() {
	int n; cin >> n;
	rep(i, n)cin >> x[i];
	kj[0] = 1;
	rep(i, n + 1)kj[i + 1] = (kj[i] * (i + 1)) % mod;
	c[0] = kj[n-1];
	rep(i, n - 2) {
		c[i + 1] = c[i] + kj[n - 1] * modpow((i + 2), mod - 2, mod) % mod;
		c[i + 1] %= mod;
	}
	int ans = 0;
	rep(i, n - 1) {
		ans += c[i] * (x[i + 1] - x[i]) % mod;
		ans %= mod;
	}
	cout << ans << endl;
}
