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 all(V) V.begin(),V.end()
#define ll long long
#define MOD 1000000007

ll factr[100010], frf[100010], frfr[100010], ans[100010];

ll modpow(ll x, ll y) {
	ll ans = 1;
	while (y) {
		if (y & 1)ans *= x;
		x *= x;
		ans %= MOD;
		x %= MOD;
		y /= 2;
	}
	return ans;
}

ll moddiv(ll x, ll y) {
	return x * modpow(y, MOD - 2) % MOD;
}

int main() {
	ll N;
	cin >> N;
	factr[0] = 1, frf[0] = 1, frfr[0] = 0;
	for (int i = 1;i < 100010;i++) {
		factr[i] = factr[i - 1] * i;
		factr[i] %= MOD;
		frf[i] = frf[i - 1] * moddiv(1, i);
		frfr[i] = (frfr[i - 1] + frf[i]) % MOD;
	}

	N--;
	ans[0] = factr[N];
	for (ll i = 1;i < N;i++) {
		ans[i] = ans[i - 1] + moddiv(factr[N], i + 1);
	}
	ll fans = 0;
	N++;
	vector<ll> X(N);
	for (int i = 0;i < N;i++) {
		cin >> X[i];
	}
	for (int i = 0;i < N - 1;i++) {
		fans += (X[i + 1] - X[i]) * ans[i];
		fans %= MOD;
	}
	cout << fans << endl;
}