Problem:
Score : 600 points

Problem Statement

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.

Subtasks

400 points will be awarded for passing the test cases satisfying N \leq 2000.

Input

Input is given from Standard Input in the following format:

N
x_1 x_2 \ldots x_N

Output

Print the answer.

Sample Input 1

3
1 2 3

Sample Output 1

5

With probability \frac{1}{2}, the leftmost slime is chosen in the first operation, in which case the total distance traveled is 2.

With probability \frac{1}{2}, the middle slime is chosen in the first operation, in which case the total distance traveled is 3.

The answer is the expected total distance traveled, 2.5, multiplied by 2!, which is 5.

Sample Input 2

12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932

Sample Output 2

750927044

Find the expected value multiplied by (N-1)!, modulo (10^9+7).

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"

using namespace std;
#ifdef _DEBUG
#define _GLIBCXX_DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif

#define int long long
#define ll long long
#define DBG 1
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define rrep(i, a, b) for (int i = (b)-1; i >= (a); i--)
#define loop(n) rep(loop, (0), (n))
#define all(c) begin(c), end(c)
const int INF = sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)(1e9) + 7;
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
using pii = pair<int, int>;
// template<class T> ostream &operator<<(ostream &os,T &t){dump(t);return os;}
template <typename T, typename S>istream &operator>>(istream &is, pair<T, S> &p) { is >> p.first >> p.second; return is; }
//template <typename T, typename S>ostream &operator<<(ostream &os, pair<T, S> &p) {os << p.first << " " << p.second;return os;}

template <typename T> void printvv(const vector<vector<T>> &v) {
	cerr << endl;
	rep(i, 0, v.size()) rep(j, 0, v[i].size()) {
		if (typeid(v[i][j]).name() == typeid(INF).name() and v[i][j] == INF) {
			cerr << "INF";
		}
		else
			cerr << v[i][j];
		cerr << (j == v[i].size() - 1 ? '\n' : ' ');
	}
	cerr << endl;
}

#ifndef _DEBUG
#define printvv(...)
#endif
void YES(bool f) { cout << (f ? "YES" : "NO") << endl; }
void Yes(bool f) { cout << (f ? "Yes" : "No") << endl; }
template <class T, class U>bool chmin(T& a, U b) { if (a > b) { a = b; return true; }return false; }
template <class T, class U>bool chmax(T& a, U b) { if (a < b) { a = b; return true; }return false; }
template <int MOD> struct ModInt {
	static const int Mod = MOD;
	unsigned x;
	ModInt() : x(0) {}
	ModInt(signed sig) {
		int sigt = sig % MOD;
		if (sigt < 0)
			sigt += MOD;
		x = sigt;
	}
	ModInt(signed long long sig) {
		int sigt = sig % MOD;
		if (sigt < 0)
			sigt += MOD;
		x = sigt;
	}
	int get() const { return (int)x; }

	ModInt &operator+=(ModInt that) {
		if ((x += that.x) >= MOD)
			x -= MOD;
		return *this;
	}
	ModInt &operator-=(ModInt that) {
		if ((x += MOD - that.x) >= MOD)
			x -= MOD;
		return *this;
	}
	ModInt &operator*=(ModInt that) {
		x = (unsigned long long)x * that.x % MOD;
		return *this;
	}
	ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }

	ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
	ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
	ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
	ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }

	ModInt inverse() const {
		signed a = x, b = MOD, u = 1, v = 0;
		while (b) {
			signed t = a / b;
			a -= t * b;
			std::swap(a, b);
			u -= t * v;
			std::swap(u, v);
		}
		if (u < 0)
			u += MOD;
		ModInt res;
		res.x = (unsigned)u;
		return res;
	}
};
template <int MOD> ostream &operator<<(ostream &os, const ModInt<MOD> &m) {
	return os << m.x;
}
template <int MOD> istream &operator>>(istream &is, ModInt<MOD> &m) {
	signed long long s;
	is >> s;
	m = ModInt<MOD>(s);
	return is;
};
// pow(mint(2), 1000) のように
template <int MOD> ModInt<MOD> pow(ModInt<MOD> a, unsigned long long k) {
	ModInt<MOD> r = 1;
	while (k) {
		if (k & 1)
			r *= a;
		a *= a;
		k >>= 1;
	}
	return r;
}

//#define int long long を使っても大丈夫
using mint = ModInt<1000000007>;

// 5は1e9+7の原始根
// return g^x=a mod p
int modlog(int g, int a) {
	// g is a primitive root modulo mint::Mod
	// let x = i*n + j , n = sqrt(Mod)
	// g^j = a*inv(g^(i*n)) = a*(inv(g^n))^i
	//https://en.wikipedia.org/wiki/Baby-step_giant-step
	//http://sonickun.hatenablog.com/entry/2016/11/20/192743
	//https://cp-algorithms.com/algebra/discrete-log.html
	unordered_map<int, int> M;
	int n = sqrt(mint::Mod);
	ll cur = 1;
	for (int i = 0; i < n; i++, (cur *= g) %= mint::Mod) {
		M[cur] = i;
	}
	ll gn = cur;
	ll gn_inv = mint(gn).inverse().get();
	cur = 1;
	ll num = 0;
	while (1) {
		ll x = a * cur % mint::Mod;
		if (M.count(x)) return num + M[x];
		cur = cur * gn_inv%mint::Mod;
		num += n;
	}
}
int modlog(int g, mint a) { return modlog(g, a.get()); }


// nCrで用いる
vector<mint> fact, factinv;
// nCrで用いる 予め計算しておく
void nCr_compute_factinv(int N) {
	N = min(N, mint::Mod - 1);
	fact.resize(N + 1);
	factinv.resize(N + 1);
	fact[0] = 1;
	rep(i, 1, N + 1) fact[i] = fact[i - 1] * i;
	factinv[N] = fact[N].inverse();
	for (int i = N; i >= 1; i--)
		factinv[i - 1] = factinv[i] * i;
}
mint nCr(int n, int r) {
	if (n >= mint::Mod)
		return nCr(n % mint::Mod, r % mint::Mod) *
		nCr(n / mint::Mod, r / mint::Mod);
	return r > n ? 0 : fact[n] * factinv[n - r] * factinv[r];
}

mint nHr(int n, int r) { return r == 0 ? 1 : nCr(n + r - 1, r); }
//http://maicommon.ciao.jp/ss/dscrtMath2/Cataran/index.htm
mint catalan(int n) { return nCr(2 * n, n) / (n + 1); }

signed main(signed argc, char *argv[]) {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(12);

	int N; cin >> N;
	if (N > 2 * 1e3)return 0;
	nCr_compute_factinv(2 * N);
	vector<int>x(N); rep(i, 0, N) { cin >> x[i]; }
	mint ans;
	rep(i, 0, N) {
		rep(j, i + 1, N) {
			dump(i, j);
			dump(j - i - 1, j - i + 1);
			dump(fact[N - 1] * fact[j - i - 1] / fact[j - i + 1 - (j == N - 1)], i, j);
			auto c = fact[N - 1] * fact[j - i - 1] / fact[j - i + 1 - (j == N - 1)];
			ans += c * (x[j] - x[i]);
		}
	}
	cout << ans << endl;

	return 0;
}