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;
using i64 = int_fast64_t;

i64 mod_pow(i64 x, i64 n, i64 MOD = 1e9 + 7){
    i64 res = 1;
    while(n > 0){
        if(n & 1) res = (res * x) % MOD ;
        x = (x * x) % MOD;
        n >>= 1;
    }
    return res;
}

i64 mod_inv(i64 x, i64 MOD = 1e9 + 7){
    return mod_pow(x, MOD - 2, MOD);
}

int main(){
    i64 N;
    cin >> N;
    vector<i64> X(N);
    for(int i = 0; i < N; i++) cin >> X[i];
    
    const i64 MOD = 1e9 + 7;
    i64 inv_s = 0;
    i64 ans = 0;
    for(int i = 0; i < N-1; i++){
        inv_s = (inv_s + mod_inv(i+1)) % MOD;
        ans = (ans + (X[i+1] - X[i]) * inv_s) % MOD;
    }
    for(int n = 1; n < N; n++) ans = (ans * n) % MOD;
    cout << ans << endl;
}