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 mod 1000000007
#define rep(i,n) for(int i=0;(i)<(n);i++)
typedef long long ll;
using namespace std;
ll fact(ll n){
    if(n<=1) return 1;
    else return ((n%mod)*fact(n-1))%mod;
}
ll modpow(ll a, ll n) {
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
ll modinv(ll a) {
    return modpow(a, mod - 2);
}
int main(void){
    ll n; cin >> n;
    ll x[100000];
    rep(i,n) cin >> x[i];
    ll p[100000]; p[0] = fact(n-1);
    ll ans = 0;
    rep(i,n-2){
        p[i+1] = (p[i] + p[0]*modinv(i+2))%mod;
        ans = (ans + ((x[i+1]-x[i]) * p[i])%mod)%mod;
    }
    ans = (ans + ((x[n-1]-x[n-2]) * p[n-2])%mod)%mod;
    cout << ans << endl;
}

