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;
const int N = 1e5 + 50;
typedef long long ll;
ll a[N];
const ll mod = 1e9 + 7;
ll pre[N];
ll fac[N];
ll qpow(ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans % mod;
}
int main() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%lld", &a[i]);
    }
    fac[0] = 1;
    for (int i = 1; i <= n - 1; i++) fac[i] = fac[i - 1] * i % mod;
    for (int i = 1; i <= n - 1; i++) {
        pre[i] = (pre[i - 1] + fac[n - 1] * qpow(i, mod - 2) % mod) % mod;
    }
    ll ans = 0;
    for (int i = 1; i <= n - 1; i++) {
        ans = (ans + pre[i] * (a[i + 1] - a[i] + mod) % mod) % mod;
    }
    printf("%lld\n", ans);
    return 0;
}