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 ll long long

ll mod = 1e9+7;

ll modpower(ll a, ll r){ //a^r
    ll x = 1;
    while (r > 0) {
        if (r & 1) x = x * a % mod;
        r >>= 1;
        a = a * a % mod;
    }
    return x;
}

ll modinv(ll a){
    return modpower(a, mod-2);
}
int main() {
    int N;
    cin >> N;
    vector<ll> x(N);
    for (int i = 0; i < N; i++) cin >> x[i];
    vector<ll> p(N+1, 0);
    vector<ll> psum(N+1, 0);
    psum[0] = 1;
    for (int i = 1; i <= N; i++) {
        p[i] = psum[i-1]*modinv(i)%mod;
        psum[i] = (psum[i-1] + p[i] + 1) % mod;
        //cout << p[i] << " " << psum[i] << endl;
    }
    ll ans = 0;
    for (int i = 1; i <= N-1; i++) {
        ans += p[i] * (x[i]-x[i-1]);
        ans %= mod;
    }
    for (int i = 2; i <= N-1; i++) {
        ans = ans * i % mod;
    }
    cout << ans << endl;
    return 0;
}