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;

typedef long long int ll;
typedef pair<ll, ll> pll;

#define FOR(i, n, m) for(ll (i)=(m);(i)<(n);++(i))
#define REP(i, n) FOR(i,n,0)
#define OF64 std::setprecision(10)

const ll MOD = 1000000007;
const ll INF = (ll) 1e15;

ll N;
ll X[100005];

ll modpow(ll n) {
    ll s = 1, p = n;
    for (ll i = 0; (1LL << i) <= MOD - 2; ++i, p = (p * p) % MOD) {
        if (((MOD - 2) & (1 << i)) == 0)
            continue;
        s *= p;
        s %= MOD;
    }
    return s;
}

int main() {
    cin >> N;
    REP(i, N) {
        cin >> X[i];
    }
    ll P = 1;
    REP(i, N - 1) {
        P = (P * (i + 1)) % MOD;
    }

    ll p = 0;
    ll ans = 0;
    REP(i, N - 1) {
        ll x = X[i + 1] - X[i];
        p = (p + modpow(i + 1) * P) % MOD;
        ans = (ans + p * x) % MOD;
    }

    cout << ans << endl;
    return 0;
}