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>
#include<vector>

const long long mod = 1000000007;

long long powMod(long long n,long long k,const long long p){
    long long res = 1;
    while(k > 0){
        if(k&1) res = res*n%p;
        k >>= 1;
        n = n*n%p;
    }
    return res;
}

int main(){
    long long n;
    std::cin >> n;
    std::vector<long long> delta(n-1,0);
    long long last;
    long long mother = 1;
    std::cin >> last;
    for(long long i=0; i<n-1; i++){
        long long x;
        std::cin >> x;
        delta.at(i) = x - last;
        last = x;
        mother = mother*(i+1)%mod;
    }
    long long total = 0, ans = 0;
    for(long long i=0; i<n-1; i++){
        long long now = (total*powMod(i+1,mod-2,mod)+1)%mod;
        ans += now*delta[i]%mod;
        ans %= mod;
        total += now;
        total %= mod;
    }
    std::cout << ans*mother%mod << "\n";
}
