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 mod 1000000007
#define rep(i,n) for(int i=0;(i)<(n);i++)
typedef long long ll;
using namespace std;
int main(void){
    int n; cin >> n;
    int inv [100003];
    int x[100003];
    rep(i,n) cin >> x[i];
    int ans = 0;
    inv[1] = 1;
    for (int i = 2; i <= n; ++i)
        inv[i] = -1ll * (mod / i) * inv[mod % i] % mod;
    int tmp = 0;
    rep(i,n-1){
        tmp = (tmp + inv[i+1])%mod;
        ans = (1ll*(x[i+1]-x[i])*tmp + ans)%mod;
    }
    rep(i,n-1) ans = 1ll*ans*(i+1)%mod;
    cout << (ans+mod)%mod << endl;
}

