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<cstdio>

using namespace std;
const int mod = 1e9 + 7;
int qs[(int)1e5 + 5], inv[(int)1e5 + 5];
int pow(int val, int idx){
    if(idx == 0) return 1;
    if(idx == 1) return val;
    int k = pow(val, idx >> 1);
    k = (1LL * k * k) %  mod;
    if(idx & 1) k = (1LL * k * val) % mod;
    return k;
}
int main(){
    int n, pre, x;
    scanf("%d %d",&n,&pre);
    for(int i = 1; i < n;i++){
        scanf("%d", &x);
        qs[i] = qs[i-1] + x - pre;
        pre = x;
    }
    int fac = 1;
    for(int i = 1;i < n;i++){
        fac = (1LL * fac * i) % mod;
        inv[i] = i == 1 ? 1 : pow(i, mod - 2);
    }
    int ans = 0;
    for(int i = 1; i < n; i++){
        ans += (1LL * (qs[n - 1] - qs[i-1]) * (1LL * fac * inv[i] % mod)) % mod;
        ans = ans >= mod ? ans - mod : ans;
    }
    printf("%d",ans);



}
