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 <cstdio>
# include <algorithm>
# include <cstring>
# include <set>
using namespace std;

long long A[100100];
long long mi[100100];
const long long mod = 1e9+7;
long long m(long long x){
    long long ans = 1;
    long long temp = x;
    long long cnt = mod-2;
    while(cnt){
        if (cnt&1) ans=ans*temp%mod;
        temp = temp*temp%mod;
        cnt>>=1;
    }
    return ans;
}
int main(void){
    long long N;
    scanf("%lld",&N);
    mi[0] = 1;
    for (long long i=1;i<=N;i++) mi[i] = mi[i-1]*i%mod;
    long long temp = mi[N-1];
    long long anss = 0;
    long long t = 0;
    scanf("%lld",&A[1]);
    for (long long i=2;i<=N;i++){
        scanf("%lld",&A[i]);
        long long tt = A[i]-A[i-1];
        t+=temp*m(i-1);
        t%=mod;
        anss+=tt*t;
        anss%=mod;
    } 
    cout<<anss<<endl;
    return 0;
}