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 <iostream>
#include <vector>

using namespace std;
typedef long long ll;

ll mod = 1e9 + 7;

vector<ll> fact;

void init_fact(int n) {
    fact = vector<ll>(n);
    fact[0] = 1;
    for (int i = 1; i < n; i++) fact[i] = fact[i - 1] * i % mod;
}

ll calc_pow(ll x, ll pow) {
    if (pow == 0) return 1;
    ll a = calc_pow(x, pow / 2);
    a = a * a % mod;
    if (pow % 2 == 1) a *= x;
    return a % mod;
}

ll calc_div(ll a, ll b) {
    return a * calc_pow(b, mod - 2) % mod;
}

int main() {
    init_fact(200000);

    int n, prev;
    cin >> n >> prev;

    ll prob = 0, ret = 0;
    for (int i = 1; i < n; i++) {
        prob += calc_div(fact[n - 1], i), prob %= mod;

        int now;
        cin >> now;
        ret += (now - prev) * prob, ret %= mod;
        prev = now;
    }

    cout << ret << endl;
    return 0;
}