Problem:
Problem E: The Two Men of the Japanese Alps

Two experienced climbers are planning a first-ever attempt: they start at two points of the equal altitudes on a mountain range, move back and forth on a single route keeping their altitudes equal, and finally meet with each other at a point on the route. A wise man told them that if a route has no point lower than the start points (of the equal altitudes) there is at least a way to achieve the attempt. This is the reason why the two climbers dare to start planning this fancy attempt.

The two climbers already obtained altimeters (devices that indicate altitude) and communication devices that are needed for keeping their altitudes equal. They also picked up a candidate route for the attempt: the route consists of consequent line segments without branches; the two starting points are at the two ends of the route; there is no point lower than the two starting points of the equal altitudes. An illustration of the route is given in Figure E.1 (this figure corresponds to the first dataset of the sample input).

Figure E.1: An illustration of a route

The attempt should be possible for the route as the wise man said. The two climbers, however, could not find a pair of move sequences to achieve the attempt, because they cannot keep their altitudes equal without a complex combination of both forward and backward moves. For example, for the route illustrated above: a climber starting at p1 (say A) moves to s, and the other climber (say B) moves from p6 to p5; then A moves back to t while B moves to p4; finally A arrives at p3 and at the same time B also arrives at p3. Things can be much more complicated and thus they asked you to write a program to find a pair of move sequences for them.

There may exist more than one possible pair of move sequences, and thus you are requested to find the pair of move sequences with the shortest length sum. Here, we measure the length along the route surface, i.e., an uphill path from (0, 0) to (3, 4) has the length of 5.

Input

The input is a sequence of datasets.

The first line of each dataset has an integer indicating the number of points N (2 ≤ N ≤ 100) on the route. Each of the following N lines has the coordinates (xi, yi) (i = 1, 2, ... , N) of the points: the two start points are (x1, y1) and (xN, yN); the line segments of the route connect (xi, yi) and (xi+1, yi+1) for i = 1, 2, ... , N - 1. Here, xi is the horizontal distance along the route from the start point x1, and yi is the altitude relative to the start point y1. All the coordinates are non-negative integers smaller than 1000, and inequality xi < xi+1 holds for i = 1, 2, .. , N - 1, and 0 = y1 = yN ≤ yi for i = 2, 3, ... , N - 1.

The end of the input is indicated by a line containing a zero.

Output

For each dataset, output the minimum sum of lengths (along the route) of move sequences until the two climbers meet with each other at a point on the route. Each output value may not have an error greater than 0.01.

Sample Input

6
0 0
3 4
9 12
17 6
21 9
33 0
5
0 0
10 0
20 0
30 0
40 0
10
0 0
1 2
3 0
6 3
9 0
11 2
13 0
15 2
16 2
18 0
7
0 0
150 997
300 1
450 999
600 2
750 998
900 0
0

Output for the Sample Input

52.5
40.0
30.3356209304689
10078.072814085803

Solution in C++11 (judge verdict: Accepted)

////////////////////////////////////////
///  tu3 pro-con template            ///
////////////////////////////////////////
#include <cassert>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <numeric>
#include <functional>
#include <vector>
#include <queue>
#include <string>
#include <complex>
#include <stack>
#include <set>
#include <map>
#include <list>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <regex>
using namespace std;

//// MACRO ////
#define countof(a) (sizeof(a)/sizeof(a[0]))

#define REP(i,n) for (int i = 0; i < (n); i++)
#define RREP(i,n) for (int i = (n)-1; i >= 0; i--)
#define FOR(i,s,n) for (int i = (s); i < (n); i++)
#define RFOR(i,s,n) for (int i = (n)-1; i >= (s); i--)
#define pos(c,i) c.being() + (i)
#define allof(c) c.begin(), c.end()
#define aallof(a) a, countof(a)
#define partof(c,i,n) c.begin() + (i), c.begin() + (i) + (n)
#define apartof(a,i,n) a + (i), a + (i) + (n)

#define EPS 1e-9
#define INF (1L << 30)
#define LINF (1LL << 60)

#define PREDICATE(t,a,exp) [&](const t & a) -> bool { return exp; }
#define COMPARISON_T(t) bool(*)(const t &, const t &)
#define COMPARISON(t,a,b,exp) [&](const t & a, const t & b) -> bool { return exp; }
#define CONVERTER(TSrc,t,TDest,exp) [&](const TSrc &t)->TDest { return exp; }

inline int sign(double x) { return (abs(x) < EPS ? 0 : x > 0 ? 1 : -1); }

//// i/o helper ////

struct _Reader { template <class T> _Reader operator ,(T &rhs) { cin >> rhs; return *this; } };
struct _Writer { bool f; _Writer() : f(false) { } template <class T> _Writer operator ,(const T &rhs) { cout << (f ? " " : "") << rhs; f = true; return *this; } };
#define READ(t,...) t __VA_ARGS__; _Reader(), __VA_ARGS__
#define WRITE(...) _Writer(), __VA_ARGS__; cout << endl

template <class T> struct vevector : public vector<vector<T>> { vevector(int n = 0, int m = 0, const T &initial = T()) : vector<vector<T>>(n, vector<T>(m, initial)) { } };
template <class T> struct vevevector : public vector<vevector<T>> { vevevector(int n = 0, int m = 0, int l = 0, const T &initial = T()) : vector<vevector<T>>(n, vevector<T>(m, l, initial)) { } };
template <class T> struct vevevevector : public vector<vevevector<T>> { vevevevector(int n = 0, int m = 0, int l = 0, int k = 0, const T &initial = T()) : vector<vevevector<T>>(n, vevevector<T>(m, l, k, initial)) { } };
template <class T> T read() { T t; cin >> t; return t; }
template <class T> vector<T> read(int n) { vector<T> v; REP(i, n) { v.push_back(read<T>()); } return v; }
template <class T> vevector<T> read(int n, int m) { vevector<T> v; REP(i, n) v.push_back(read<T>(m)); return v; }
template <class T> vevector<T> readjag() { return read<T>(read<int>()); }
template <class T> vevector<T> readjag(int n) { vevector<T> v; REP(i, n) v.push_back(readjag<T>()); return v; }

template <class T1, class T2> inline istream & operator >> (istream & in, pair<T1, T2> &p) { in >> p.first >> p.second; return in; }
template <class T1, class T2> inline ostream & operator << (ostream &out, pair<T1, T2> &p) { out << p.first << p.second; return out; }
template <class T> inline ostream & operator << (ostream &out, vector<T> &v)
{
	ostringstream ss;
	for (auto x : v) ss << x << ' ';
	auto s = ss.str();
	out << s.substr(0, s.length() - 1) << endl;
	return out;
}

//// start up ////
void solve();
int main()
{
	solve();
	return 0;
}

//// geometry ////
/// 2次元
struct P2
{
	double x, y;
	P2(double x = 0, double y = 0) : x(x), y(y) { }
	P2(complex<double> c) : x(c.real()), y(c.imag()) { }
	P2 operator +() const { return *this; }
	P2 operator +(const P2 &_) const { return P2(x + _.x, y + _.y); }
	P2 operator -() const { return P2(-x, -y); }
	P2 operator -(const P2 &_) const { return *this + -_; }
	P2 operator *(double _) const { return P2(x*_, y*_); }
	P2 operator /(double _) const { return P2(x / _, y / _); }
	double dot(const P2 &_) const { return x*_.x + y*_.y; } // 内積
	double cross(const P2 &_) const { return x*_.y - y*_.x; } // 外積
	double sqlength() const { return x*x + y*y; } // 二乗長さ
	double length() const { return sqrt(sqlength()); } // 長さ
	P2 orthogonal() const { return P2(y, -x); }
	P2 direction() const { return *this / length(); } // 方向ベクトル
};
inline istream & operator>>(istream & in, P2 & p) { in >> p.x >> p.y; return in; }
inline ostream & operator<<(ostream & out, const P2 & p) { out << p.x << ' ' << p.y; return out; }
inline double abs(P2 p2) { return p2.length(); }

// 範囲内チェック
inline bool contains(int x, int min, int max) { return x >= min && x < max; }
inline bool contains(int x, int y, int W, int H) { return x >= 0 && x < W && y >= 0 && y < H; }

////////////////////
/// template end ///
////////////////////

void solve()
{
	int cases = INF;
	REP(_, cases)
	{
		READ(int, n);
		if (!n) { break; }
		auto ps = read<P2>(n);
		
		vector<P2> g;
		for (auto p : ps)
		{
			g.emplace_back(p.x, p.y);
			REP(j, n - 1)
			{
				P2 p1 = ps[j], p2 = ps[j + 1];

				if (min(p1.y, p2.y) < p.y && max(p1.y, p2.y) > p.y)
				{
					P2 d = p - p1, d2 = p2 - p1;
					g.emplace_back(d.y / d2.y * d2.x + p1.x, p.y);
				}
			}
		}
		sort(allof(g), COMPARISON(P2, a, b, a.x < b.x));

		struct S { pair<int, int> next; double cost; };
		double result = 0;

		priority_queue<S, vector<S>, COMPARISON_T(S)> q(COMPARISON(S, a, b, a.cost > b.cost));
		q.push({ { 0, g.size() - 1 }, 0 });
		set<pair<int, int>> visited;
		while (!q.empty())
		{
			S c = q.top(); q.pop();
			if (visited.count(c.next)) continue;
			visited.insert(c.next);
			if (c.next.first == c.next.second) { result = c.cost; break; }

			int p1 = c.next.first, p2 = c.next.second;
			int dp[] = { -1, 0, 1 };
			REP(i, 3) REP(j, 3)
			{
				int np1 = p1 + dp[i], np2 = p2 + dp[j];
				if (!contains(np1, np2, g.size(), g.size())) continue;
				if (visited.count({ np1, np2 })) continue;
				if (abs(g[np1].y - g[np2].y) > EPS) continue;
				double cost = (g[np1] - g[p1]).length() + (g[np2] - g[p2]).length();
				q.push({ { np1, np2 }, c.cost + cost });
			}
		}
		printf("%.16f\n", result);
	}
}