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>
const long long mod = 1e9+7;
long long x[100005];

long long pow(long long a,long long b){
    long long res = 1;
    while(b){
        if(b&1) res = res*a%mod;
        a = a*a%mod;
        b /= 2;
    }
    return res;
}

long long inv(long long u){ return pow(u,mod-2); }

int main(){
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++) scanf("%lld",&x[i]);
    long long ans = 0;
    long long count = 0;
    long long hold = 1;
    for(int i = 2; i <= n-1; i++) hold = hold*(long long)i%mod;
    //printf("hold = %lld\n",hold);
    for(int i = 1; i <= n-1; i++){
        count = (count + inv(i))%mod;
        //printf("i = %d, time = %lld\n",i,count*hold%mod);
        ans = (ans + count*(x[i+1]-x[i])%mod)%mod;
    }
    printf("%lld\n",ans*hold%mod);
    return 0;
}


