Problem: You are playing a game and your goal is to maximize your expected gain. At the beginning of the game, a pawn is put, uniformly at random, at a position p\in\\{1,2,\dots, N\\}. The N positions are arranged on a circle (so that 1 is between N and 2).

The game consists of turns. At each turn you can either end the game, and get A_p dollars (where p is the current position of the pawn), or pay B_p dollar to keep playing. If you decide to keep playing, the pawn is randomly moved to one of the two adjacent positions p-1, p+1 (with the identifications 0 = N and N+1=1).

What is the expected gain of an optimal strategy?

Note: The "expected gain of an optimal strategy" shall be defined as the supremum of the expected gain among all strategies such that the game ends in a finite number of turns.

Constraints

* 2 \le N \le 200,000
* 0 \le A_p \le 10^{12} for any p = 1,\ldots, N
* 0 \le B_p \le 100 for any p = 1, \ldots, N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 \cdots A_N
B_1 B_2 \cdots B_N


Output

Print a single real number, the expected gain of an optimal strategy. Your answer will be considered correct if its relative or absolute error does not exceed 10^{-10}.

Examples

Input

5
4 2 6 3 5
1 1 1 1 1


Output

4.700000000000


Input

4
100 0 100 0
0 100 0 100


Output

50.000000000000


Input

14
4839 5400 6231 5800 6001 5200 6350 7133 7986 8012 7537 7013 6477 5912
34 54 61 32 52 61 21 43 65 12 45 21 1 4


Output

7047.142857142857


Input

10
470606482521 533212137322 116718867454 746976621474 457112271419 815899162072 641324977314 88281100571 9231169966 455007126951
26 83 30 59 100 88 84 91 54 61


Output

815899161079.400024414062

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>
#include<set>
#include<queue>
#include<map>
#include<algorithm>
#include<cstring>
#include<string>
#include<cassert>
#include<cmath>
#include<climits>
#include<iomanip>
#include<stack>
#include<unordered_map>
#include<bitset>
#include<limits>
#include<complex>
#include<array>
#include<numeric>
#include<functional>

using namespace std;
#define ll long long
#define ull unsigned long long
#define rrep(i,m,n) for(ll (i)=(ll)(m);(i)>=(ll)(n);(i)--)
#define rep(i,m,n) for(ll (i)=(ll)(m);i<(ll)(n);i++)
#define REP(i,n) rep(i,0,n)
#define FOR(i,c) for(decltype((c).begin())i=(c).begin();i!=(c).end();++i)
#define all(hoge) (hoge).begin(),(hoge).end()
typedef pair<ll, ll> P;
constexpr long double m_pi = 3.1415926535897932L;
constexpr ll MOD = 1000000007;
constexpr ll INF = 1LL << 61;
constexpr long double EPS = 1e-10;
template<typename T> using vector2 = vector<vector<T>>;
template<typename T> using vector3 = vector<vector2<T>>;
typedef vector<ll> Array;
typedef vector<Array> Matrix;
string operator*(const string& s, int k) {
	if (k == 0) return "";
	string p = (s + s) * (k / 2);
	if (k % 2 == 1) p += s;
	return p;
}

template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

struct Edge {//グラフ
	int to, rev; ll cap;
	Edge(int _to, ll _cap, int _rev) {
		to = _to; cap = _cap; rev = _rev;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;
void add_edge(Graph& G, int from, int to, ll cap, bool revFlag, ll revCap) {//最大フロー求める Ford-fulkerson
	G[from].push_back(Edge(to, cap, (ll)G[to].size()));
	if (revFlag)G[to].push_back(Edge(from, revCap, (ll)G[from].size() - 1));//最小カットの場合逆辺は0にする
}

ll max_flow_dfs(Graph& G, ll v, ll t, ll f, vector<bool>& used)
{
	if (v == t)
		return f;
	used[v] = true;
	for (int i = 0; i < G[v].size(); ++i) {
		Edge& e = G[v][i];
		if (!used[e.to] && e.cap > 0) {
			ll d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
			if (d > 0) {
				e.cap -= d;
				G[e.to][e.rev].cap += d;
				return d;
			}
		}
	}
	return 0;
}
//二分グラフの最大マッチングを求めたりも出来る　また二部グラフの最大独立集合は頂点数-最大マッチングのサイズ
ll max_flow(Graph& G, ll s, ll t)//O(V(V+E))
{
	ll flow = 0;
	for (;;) {
		vector<bool> used(G.size());
		REP(i, used.size())used[i] = false;
		ll f = max_flow_dfs(G, s, t, INF, used);
		if (f == 0) {
			return flow;
		}
		flow += f;
	}
}

void BellmanFord(Graph& G, ll s, Array& d, Array& negative) {//O(|E||V|)
	d.resize(G.size());
	negative.resize(G.size());
	REP(i, d.size())d[i] = INF;
	REP(i, d.size())negative[i] = false;
	d[s] = 0;
	REP(k, G.size() - 1) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
				}
			}
		}
	}
	REP(k, G.size() - 1) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
					negative[G[i][j].to] = true;
				}
				if (negative[i] == true)negative[G[i][j].to] = true;
			}
		}
	}
}
void Dijkstra(Graph& G, ll s, Array& d) {//O(|E|log|V|)
	d.resize(G.size());
	REP(i, d.size())d[i] = INF;
	d[s] = 0;
	priority_queue<P, vector<P>, greater<P>> q;
	q.push(make_pair(0, s));
	while (!q.empty()) {
		P a = q.top();
		q.pop();
		if (d[a.second] < a.first)continue;
		REP(i, G[a.second].size()) {
			Edge e = G[a.second][i];
			if (d[e.to] > d[a.second] + e.cap) {
				d[e.to] = d[a.second] + e.cap;
				q.push(make_pair(d[e.to], e.to));
			}
		}
	}
}
void WarshallFloyd(Graph& G, Matrix& d) {//O(V^3)
	d.resize(G.size());
	REP(i, d.size())d[i].resize(G.size());
	REP(i, d.size()) {
		REP(j, d[i].size()) {
			d[i][j] = ((i != j) ? INF : 0);
		}
	}
	REP(i, G.size()) {
		REP(j, G[i].size()) {
			chmin(d[i][G[i][j].to], G[i][j].cap);
		}
	}
	REP(i, G.size()) {
		REP(j, G.size()) {
			REP(k, G.size()) {
				chmin(d[j][k], d[j][i] + d[i][k]);
			}
		}
	}
}
bool tsort(Graph& graph, Array& order) {//トポロジカルソートO(E+V)
	int n = graph.size(), k = 0;
	Array in(n);
	for (auto& es : graph)
		for (auto& e : es)in[e.to]++;
	priority_queue<ll, Array, greater<ll>> que;
	REP(i, n)
		if (in[i] == 0)que.push(i);
	while (que.size()) {
		int v = que.top();
		que.pop();
		order.push_back(v);
		for (auto& e : graph[v])
			if (--in[e.to] == 0)que.push(e.to);
	}
	if (order.size() != n)return false;
	else return true;
}
class Lca {
public:
	const int n = 0;
	const int log2_n = 0;
	std::vector<std::vector<int>> parent;
	std::vector<int> depth;

	Lca() {}

	Lca(const Graph& g, int root)
		: n(g.size()), log2_n(log2(n) + 1), parent(log2_n, std::vector<int>(n)), depth(n) {
		dfs(g, root, -1, 0);
		for (int k = 0; k + 1 < log2_n; k++) {
			for (int v = 0; v < (int)g.size(); v++) {
				if (parent[k][v] < 0)
					parent[k + 1][v] = -1;
				else
					parent[k + 1][v] = parent[k][parent[k][v]];
			}
		}
	}

	void dfs(const Graph& g, int v, int p, int d) {
		parent[0][v] = p;
		depth[v] = d;
		for (auto& e : g[v]) {
			if (e.to != p) dfs(g, e.to, v, d + 1);
		}
	}

	int get(int u, int v) {
		if (depth[u] > depth[v]) std::swap(u, v);
		for (int k = 0; k < log2_n; k++) {
			if ((depth[v] - depth[u]) >> k & 1) {
				v = parent[k][v];
			}
		}
		if (u == v) return u;
		for (int k = log2_n - 1; k >= 0; k--) {
			if (parent[k][u] != parent[k][v]) {
				u = parent[k][u];
				v = parent[k][v];
			}
		}
		return parent[0][u];
	}
};

class UnionFind {
	vector<int> data;
	ll num;
public:
	UnionFind(int size) : data(size, -1), num(size) { }
	bool unite(int x, int y) {//xとyの集合を統合する
		x = root(x); y = root(y);
		if (x != y) {
			if (data[y] < data[x]) swap(x, y);
			data[x] += data[y]; data[y] = x;
		}
		num -= (x != y);
		return x != y;
	}
	bool findSet(int x, int y) {//xとyが同じ集合か返す
		return root(x) == root(y);
	}
	int root(int x) {//xのルートを返す
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	ll size(int x) {//xの集合のサイズを返す
		return -data[root(x)];
	}
	ll numSet() {//集合の数を返す
		return num;
	}
};

template<typename T, typename F>
class SegmentTree {
private:
	T identity;
	F merge;
	ll n;
	vector<T> dat;
public:
	SegmentTree(F f, T id,vector<T> v) :merge(f), identity(id) {
		int _n = v.size();
		n = 1;
		while (n < _n)n *= 2;
		dat.resize(2 * n - 1, identity);
		REP(i, _n)dat[n + i - 1] = v[i];
		for (int i = n - 2; i >= 0; i--)dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]);
	}
	SegmentTree(F f, T id, int _n) :merge(f), identity(id) {
		n = 1;
		while (n < _n)n *= 2;
		dat.resize(2 * n - 1, identity);
	}
	void set_val(int i, T x) {
		i += n - 1;
		dat[i] = x;
		while (i > 0) {
			i = (i - 1) / 2;
			dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]);
		}
	}
	T query(int l, int r) {
		T left = identity, right = identity;
		l += n - 1; r += n - 1;
		while (l < r) {
			if ((l & 1) == 0)left = merge(left, dat[l]);
			if ((r & 1) == 0)right = merge(dat[r - 1], right);
			l = l / 2;
			r = (r - 1) / 2;
		}
		return merge(left, right);
	}
};

template< typename T >
class FenwickTree {
	vector< T > data;
	int n;
	int p;
public:
	FenwickTree(int n) :n(n) { 
		data.resize(n + 1LL, 0);
		p = 1;
		while (p < data.size())p *= 2;
	}
	T sum(int k) {
		T ret = 0;
		for (; k > 0; k -= k & -k) ret += data[k];
		return (ret);
	}
	
	T sum(int a, int b) { return sum(b) - sum(a); }//[a,b)

	void add(int k, T x) {
		for (++k; k <= n; k += k & -k) data[k] += x;
	}

	int lower_bound(ll w) {
		if (w <= 0)return -1;
		int x = 0;
		for (int k = p / 2; k > 0; k /= 2) {
			if (x + k <= n && data[x + k] < w)w -= data[x + k], x += k;
		}
		return x;
	}
};



//約数求める //約数
void divisor(ll n, vector<ll>& ret) {
	for (ll i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(ret.begin(), ret.end());
}

void prime_factorization(ll n, vector<P>& ret) {
	for (ll i = 2; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back({ i,0 });
			while (n % i == 0) {
				n /= i;
				ret[ret.size() - 1].second++;
			}
		}
	}
	if (n != 1)ret.push_back({ n,1 });
}


ll mod_pow(ll x, ll n, ll mod) {
	ll res = 1;
	while (n > 0) {
		if (n & 1) res = res * x % mod;
		x = x * x % mod;
		n >>= 1;
	}
	return res;
}

ll mod_inv(ll x, ll mod) {
	return mod_pow(x, mod - 2, mod);
}

class Combination {
public:
	Array fact;
	Array inv;
	ll mod;
	ll mod_inv(ll x) {
		ll n = mod - 2LL;
		ll res = 1LL;
		while (n > 0) {
			if (n & 1) res = res * x % mod;
			x = x * x % mod;
			n >>= 1;
		}
		return res;
	}
	//if n >= mod use lucas 
	ll nCr(ll n, ll r) {
		if (n < r)return 0;
		if (n < mod)return ((fact[n] * inv[r] % mod) * inv[n - r]) % mod;

		ll ret = 1;
		while (n || r) {
			ll _n = n % mod, _r = r % mod;
			n /= mod; r /= mod;
			(ret *= nCr(_n, _r)) %= mod;
		}
		return ret;
	}
	ll nPr(ll n, ll r) {
		return (fact[n] * inv[n - r]) % mod;
	}
	ll nHr(ll n, ll r) {
		return nCr(r + n - 1, r);
	}
	Combination(ll _n, ll _mod) {
		mod = _mod;
		ll n = min(_n + 1, mod);
		fact.resize(n);
		fact[0] = 1;
		REP(i, n - 1) {
			fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
		}
		inv.resize(n);
		inv[n - 1] = mod_inv(fact[n - 1]);
		for (int i = n - 1; i > 0; i--) {
			inv[i - 1] = inv[i] * i % mod;
		}
	}
};

ll popcount(ll x) {
	x = (x & 0x5555555555555555) + (x >> 1 & 0x5555555555555555);
	x = (x & 0x3333333333333333) + (x >> 2 & 0x3333333333333333);
	x = (x & 0x0F0F0F0F0F0F0F0F) + (x >> 4 & 0x0F0F0F0F0F0F0F0F);
	x = (x & 0x00FF00FF00FF00FF) + (x >> 8 & 0x00FF00FF00FF00FF);
	x = (x & 0x0000FFFF0000FFFF) + (x >> 16 & 0x0000FFFF0000FFFF);
	x = (x & 0x00000000FFFFFFFF) + (x >> 32 & 0x00000000FFFFFFFF);

	return x;
}



long double c[202020];
long double f[202020];

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	ll n;
	cin >> n;
	vector<long double> a(n), b(n);
	REP(i, n)cin >> a[i];
	REP(i, n)cin >> b[i];
	ll t = max_element(all(a)) - a.begin();
	rotate(a.begin(), a.begin() + t, a.end());
	rotate(b.begin(), b.begin() + t, b.end());
	a.push_back(a[0]);
	rep(i, 2, n+1)c[i] = 2.0 * (c[i - 1] + b[i - 1]) - c[i - 2];
	using T = pair<long double, ll >;
	vector<T> convex;
	auto grad = [](T a, T b) {return (a.first - b.first) /(long double) (a.second - b.second);};
	REP(i, a.size()) {
		while (convex.size()>1) {
			long double pre = grad(convex.back(), convex[convex.size() - 2]);
			long double now = grad({ a[i] - c[i] ,i }, convex.back());
			if (now >= pre)convex.pop_back();
			else break;
		}
		convex.push_back({ a[i] - c[i],i });
	}
	REP(i, convex.size() - 1) {
		auto l = convex[i];
		auto r = convex[i + 1];
		auto g = grad(r, l);
		for (ll j = l.second; j <= r.second; j++) f[j] = l.first + g * (long double)(j - l.second);
	}
	long double ans = 0;
	REP(i, n)ans += f[i] + c[i];
	cout << fixed << setprecision(15) << ans / (long double)n << "\n";




	return 0;
}