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>
#define rep(i, n) for(int i=0; i<(n); i++)
#define itrep(type, it, t) for(type::iterator it=t.begin(); it!=t.end(); it++)
using namespace std;

#define MOD (1000000007)

long long invmodp(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long 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(void) {

    int N;
    cin >> N;
    long long X[N];
    long long fact = 1;
    rep(i, N) {
        cin >> X[i];
        fact = fact * max(1, i) % MOD;
    }

    long long ans = 0;
    long long inv = 0;
    rep(i, N-1) {
        long long diff = X[i+1] - X[i];
        inv = inv + invmodp(i+1, MOD) % MOD;
        long long add = diff * inv % MOD;
        ans = (ans + add) % MOD;
    }

    cout << (ans*fact)%MOD;
    return 0;
}