#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<iomanip>
const double eps=1e-3;
using namespace std;
typedef long long LL;
#define int long 
#define double long double
template <typename T>
inline void read(T &x)
{
    T f = 1;
    x = 0;
    char ch = getchar();
    while (0 == isdigit(ch))
    {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (0 != isdigit(ch))
        x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
    x *= f;
}
template <typename T>
inline void write(T x)
{
    if (x < 0)
    {
        x = ~(x - 1);
        putchar('-');
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
int dcmp(double x)
{
	if(fabs(x) < eps) return 0;
	else return x < 0 ? -1 : 1;
}
struct point
{
	double x, y;
	point (double _x = 0, double _y = 0) : x(_x) , y(_y){}
	point rot(const double rad) const { return {x * cos(rad) - y * sin(rad), x * sin(rad) + y * cos(rad)}; }
};
point operator + (point a , point b) {return point(a.x + b.x,a.y + b.y);}
point operator - (point a , point b) {return point(a.x - b.x , a.y - b.y);}
point operator * (point a , double t) {return point(a.x*t,a.y*t);}
point operator / (point a , double t) {return point(a.x/t,a.y/t);}
bool operator == (const point &a , const point b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;}
bool operator <(const point& A,const point& B){
	if(A.x == B.x) return A.y < B.y;
	return A.x < B.x;
}
double cross (point a , point b) {return a.x * b.y - b.x * a.y;}
double dot (point a , point b) {return a.x * b.x + a.y * b.y;}
double len(point v) {return sqrt(dot(v,v));}
struct Line
{
	point p,v;
	Line(){}//reserve需要
	Line(point _p , point _v) : p(_v) , v(_p){}
	point get_point_in_line(double t)
	{
		return v + (p - v) * t;
	}
	double dis(const point &a) const {return abs(cross(v - p , a - p)) / len(v - p); }
};

//思路：先找到大于s的线段，标记为a线段，第一个线段标记为b线段
//af，bf表示对于这个线段来说，从这个线段起点到af/bf的距离

void solve()
{
	double s;
	int n;
	cin >> s >> n;
	vector<point> p(n);
	for(int i = 0 ; i < n ; i ++)
	{
		cin >> p[i].x >> p[i].y;
	}
	auto res = [&](int st , double offset)
	{
		return len(p[st + 1] - p[st]) - offset;
	};
	auto getp = [&](int st , double offset)
	{
		double lenth = len(p[st + 1] - p[st]);
		return p[st] + (p[st + 1] - p[st]) * offset/lenth;
	};
	int as = 0 , bs = 0;//表示所处与第几个条线段
	double af = 0.0L , bf = 0.0L;//表示最后一段线段所走的长度
	while(as < n - 1)
	{
		double r = res(as,af);
		if(s > r)
		{
			s -= r;
			as++;
			af = 0;
		}else{
			af = s;
			break;
		}
	}

	double ans = 1e18;
	while(as < n - 1)
	{
		double ar = res(as,af),br = res(bs,bf);//a能走最远和b能走的最远
		double l = 0 , r = min(ar,br);
		while(r - l > 1e-8)
		{
			double m1 = (l + l + r)/3 , m2 = (l + r + r)/3;
			point a1 = getp(as,af+m1),a2 = getp(as,af+m2);
			point b1 = getp(bs,bf+m1),b2 = getp(bs,bf+m2);
			if(len(a1-b1) < len(a2-b2)) r = m2;
			else l = m1;
		}
		point pa = getp(as,af+l) , pb = getp(bs,bf+l);
		ans = min(ans,len(pa-pb));
		if(ar < br)
		{
			bf += ar;
			as ++;
			af = 0;
		}else{
			af += br;
			bs ++;
			bf = 0;
		}
	}
	cout << fixed << setprecision(10) << ans << char(10);
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int t = 1;
	// cin >> t;
	while(t --)
	{	
		solve();
	}
	return 0;
}