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;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (n); i++)

const ll MOD = 1e9 + 7;

ll factorial(ll n, ll m = 2) {
    // calculate n!
    m = max(2LL, m);
    ll rtn = 1;
    for (ll i = m; i <= n; i++) {
        rtn = (rtn * i) % MOD;
    }
    return rtn;
}

ll modinv(ll a, ll m) {
    ll b = m, u = 1, v = 0;
    while (b) {
        ll t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0)
        u += m;
    return u;
}

int main() {
    int n;
    cin >> n;
    vector<ll> x(n);
    rep(i, n) cin >> x[i];

    ll ans = 0;
    ll factor = factorial(n - 1);
    ll running = factor;
    for (int i = 0; i < n - 1; i++) {
        ll d = x[i + 1] - x[i];
        ans += (running % MOD) * (d % MOD);
        ans %= MOD;
        running += (factor * modinv(i + 2, MOD)) % MOD;
    }
    cout << ans << endl;
}