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 <iostream>
using namespace std;
typedef long long ll;

const ll MOD = 1000000007;

ll modpow(ll x, ll n){
    ll res = 1;
    ll mul = x;
    while(n){
        if(n % 2) res = res * mul % MOD;
        mul = mul * mul % MOD;
        n /= 2;
    }
    return res;
}

int main()
{
    ll n;
    cin >> n;
    ll x[100002];
    for(int i = 0; i < n; i++) cin >> x[i];
    ll ans = 0;
    ll r = 0;
    for(int i = 1; i < n; i++){
        r = (r + modpow(i, MOD - 2)) % MOD;
        ans = (ans + r * (x[i] - x[i - 1]) % MOD) % MOD;
    }
    for(int i = 1; i < n; i++) ans = (ans * i) % MOD;
    cout << ans << endl;
}