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>

#define ll long long
#define fornum(A,B,C) for(A=B;A<C;++A)
#define pii pair<int,int>
#define pll pair<ll,ll>

using namespace std;

/////////////////////////////////////////////////////
#define MOD (ll)(1e9 + 7)

ll moddiv(ll a,ll b){
    a %= MOD;
    b %= MOD;
    for(ll m=MOD-2;m>0;m/=2){
        if(m&1){
            a=a*b%MOD;
        }
        b=b*b%MOD;
    }
    return a;
}

ll N;
ll X[101010];
ll Fa[101010];
ll i, j, ans;
int main(){
    scanf("%lld", &N);
    fornum(i,0,N){
        scanf("%lld", &X[i]);
    }

    Fa[0] = 1;
    fornum(i,0,N){
        Fa[i + 1] = Fa[i] * (i + 1) % MOD;
    }

    j = 0;
    fornum(i,1,N){
        j += moddiv(Fa[N - 1], Fa[i]) * Fa[i - 1];
        j %= MOD;
        ans = (ans + j * (X[i] - X[i - 1])) % MOD;
    }
    printf("%lld", ans);

    return 0;
}