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++)
#define all(a) (a).begin(), (a).end()
#define pb push_back

using ll = long long;
using i_i = tuple<int, int>;

const int MOD = 1e9 + 7;

int pow_mod(ll x, ll n, int M) {
    ll ans = 1;
    for (; n; n>>=1) {
        if (n & 1) ans = ans * x % M;
        x = x * x % M;
    }
    return ans;
}

int inv_mod(int x, int p) {
    return pow_mod(x, p - 2, p);
}

int main() {
    int N; cin >> N;
    vector<int> x(N);
    rep(i, N) cin >> x[i];
    vector<ll> a(N - 1);
    rep(i, N - 1) a[i] = x[i + 1] - x[i];
    N--;
    ll c = 0, ans = 0;
    rep(i, N) {
        c = (c + inv_mod(i + 1, MOD)) % MOD;
        ans = (ans + c * a[i]) % MOD;
    }
    for (int x = 1; x <= N; x++) ans = ans * x % MOD;
    cout << ans << endl;
}
