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)

typedef long long ll;
typedef pair<int, int> P;

const int MAX = 1e5 + 5;
const int INF = 1001001001;
const int MOD = 1e9 + 7;

int main(){
    int N;
    cin >> N;

    vector<int> X(N);
    rep(i, N) cin >> X[i];

    ll fact = 1;
    for (int i = 1; i < N; ++i) {
        fact *= i;
        fact %= MOD;
    }

    vector<ll> inv(N);
    inv[1] = 1;
    for (int i = 2; i < N; ++i) {
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
    }

    vector<ll> V(N - 1);
    V[0] = 1;
    for (int i = 1; i < N - 1; ++i) {
        V[i] = V[i - 1] + 1 * inv[i + 1];
        V[i] %= MOD;
    }

    ll res = 0;
    rep(i, N - 1) {
        res += (X[i + 1] - X[i]) * V[i];
        res %= MOD;
    }

    ll ans = res * fact;
    ans %= MOD;

    cout << ans << endl;
}