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
#define all(aaa) aaa.begin(), aaa.end()

const int N = 1e5 + 5, MOD = 1e9 + 7;
int a[N];

ll bp(ll a, ll p) {
    ll r = 1;
    while (p > 0) {
        if (p & 1)
            r = r * a % MOD;
        a = a * a % MOD;
        p >>= 1;
    }
    return r;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n;
    cin >> n;

    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }

    ll ans = 0, x = 0;

    for (int i = 0; i + 1 < n; i++) {
        x = (x + bp(i + 1, MOD - 2)) % MOD;
        ans = (ans + x * (a[i + 1] - a[i])) % MOD;
    }

    for (int i = 1; i <= n - 1; i++) {
        ans = ans * i % MOD;
    }

    cout << ans;

    return 0;
}