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 <algorithm>
#include <cstdio>
#include <vector>
 
using namespace std;
 
#define REP(i, n) for(int i = 0; i < (int)(n); i++)
 
int getInt() {
    int s;
    scanf("%d", &s);
    return s;
}
 
typedef long long ll;
const ll mod = 1000000007;
 
ll x[100000];
ll f[100000 + 10];
ll rf[100000 + 10];
 
ll mpow(ll b, ll e) {
    ll ans = 1;
    while(e){
        if(e % 2 == 1) ans = (ans * b) % mod;
        b = (b * b) % mod;
        e = e / 2;
    }
    return ans;
}
 
int main(){
    const int n = getInt();
    REP(i,n) x[i] = getInt();
 
    f[0] = 1;
    REP(i,n) f[i + 1] = f[i] * (i + 1) % mod;

    ll ans = 0;
    ll acc = 0;
    for(int i = 1; i < n; i++) {
        acc = (acc + mpow(i, mod - 2)) % mod;
        ans = (ans + (x[i] - x[i - 1]) * acc % mod) % mod;
    }
 
    ans = ans * f[n - 1] % mod;
 
    printf("%lld\n", ans);
 
    return 0;
}