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;
#define rep(i,n) for (int i = 0; i < (n); ++i)
typedef long long ll;
const long long INF = 1e18;
const int MOD = 1000000007;

long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

ll fact(ll k){
    ll res=1;
    rep(i,k){
        res*=i+1;
        res%=MOD;
    }
    return res;
}


int main(){
    ll N;
    ll res=0;
    cin>>N;
    ll fac=fact(N-1);
    vector<ll> x(N);
    rep(i,N) cin>>x[i];
    rep(i,N-1){
        res+=((x[N-1]-x[i])*fac % MOD) * modinv(i+1,MOD);
        res%=MOD;
    }
    cout<<res<<endl;
    return 0;
}