#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<cmath>
#include<iomanip>
using namespace std;
// #define debug(x) cout<<"[debug]"#x<<"="<<x<<endl
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
const int INF=0x3f3f3f3f,mod=2147483647;
const ld eps=1e-7;

#define ONLINE_JUDGE
#ifndef ONLINE_JUDGE
#define debug(...)
#include<debug>
#else
#define debug(...)
#endif

const ld pi=acos(-1);
using _T=long double;

template<typename T> struct point
{
    T x,y;

    bool operator ==(const point &a) const {return abs(x-a.x)<eps&&abs(y-a.y)<eps;}
    point operator +(const point &a) const {return {x+a.x,y+a.y};}
    point operator -(const point &a) const {return {x-a.x,y-a.y};}
    point operator -() const{return {-x,-y};}
    point operator *(const T k) const {return {k*x,k*y};}
    point operator /(const T k) const {return {x/k,y/k};}
    T operator *(const point &a) const {return x*a.x+y*a.y;}//Dot
    T operator ^(const point &a) const {return x*a.y-a.x*y;}//Cross
    bool operator <(const point &a) const {if(abs(x-a.x)<eps) return y<a.y-eps; return x<a.x-eps;}
    bool is_par(const point &a) const {return abs((*this)^a)<=eps;}
    bool is_ver(const point &a) const {return abs((*this)*a)<=eps;}
    int toleft (const point &a) const {auto t=(*this)^a; return (t>eps)-(t<-eps);}//在左侧为1
    T len2() const {return (*this)*(*this);}
    T dis2(const point &a) const {return (a-(*this)).len2();}
    long double len() const {return sqrt(len2());}//点到原点的距离
    long double dis(const point &a) const {return (a-(*this)).len();};//两点之间的距离
    long double ang(const point &a) const {return acos(((*this)*a)/(this->len()*a.len()));}//两个向量的夹角
    point rot(const long double rad) const {return {x*cos(rad)-y*sin(rad),x*sin(rad)+y*cos(rad)};}// 逆时针旋转（给定角度）
    point rot(const long double cosr,const long double sinr) const {return {x*cosr-y*sinr,x*sinr+y*cosr};}// 逆时针旋转（给定角度的正弦与余弦）
};

using Point=point<_T>;

template<typename T> struct line
{
    point<T> p,v;//p+vt

    bool operator ==(const line &a) const {return v.toleft(a.v)==0&&v.toleft(p-a.p)==0;}//两条直线是否重合
    bool is_par(const line &a) const {return (v.is_par(a.v)&&!v.is_par(p-a.p));}//平行且不重合为true
    bool is_ver(const line &a) const {return v.is_ver(a.v);}//两条直线是否垂直
    bool is_on(const point<T> &a) const {return v.is_par(a-p);}//点是否在直线上
    int toleft(const point<T> &a) const {return v.toleft(a-p);}//点在直线向量方向的左侧为1，右侧为-1，直线上为0
    point<T> inter(const line &a) const {return p+v*((a.v^(p-a.p))/(v^a.v));}//求两条直线的交点
    long double dis(const point<T> &a) const {return abs(v^(a-p)/v.len());}//点到直线的距离
    long double get_rad() const {return atan2(v.y,v.x);}
    point<T> proj(const point<T> &a) const {return p+v*((v*(a-p))/(v*v));}//点在向量v上的投影坐标
    long double proj_len(const point<T> &a) const {return (v*a)/v.len();}//向量在向量v上的投影长度
};

using Line=line<_T>;
struct Circle
{
    Point o;//圆心
    ld r;//半径
};
int dcmp(ld x,ld y)
{
    if(fabs(x-y)<eps)
    return 0;
    else if(x<y) return -1;
    else return 1;
}
Circle get_circle(Point a,Point b,Point c)//三点确定一个圆
{
    Line l1={ (a+b)/2 , (a-b).rot(pi/2)};
    Line l2={ (a+c)/2 , (a-c).rot(pi/2)};
    Point o=l1.inter(l2);//此为圆心
    return {o,o.dis(a)};
}
const int N=2003;
ld dp[N][N];
int n,k;
Point p[N];
int x[N],y[N];
ld w[N][N];
int main()
{
    cerr<<fixed<<setprecision(2);

    scanf("%d%d%d",&n,&k,&x[1]);
    y[1]=(233811181ll*x[1]+1)%mod;
    for(int i=2;i<=n;i++)
    {
        x[i]=(233811181ll*y[i-1]+1)%mod;
        y[i]=(233811181ll*x[i]+1)%mod;
    }
    for(int i=1;i<=n;i++)
    {
        p[i]={(ld)x[i],(ld)y[i]};
    }
    for(int fir=1;fir<=n;fir++)
    {
        Circle c={p[fir],0};
        for(int i=fir+1;i<=n;i++)
        {
            if(dcmp(c.o.dis(p[i]),c.r)>0)
            {
                c={p[i],0};
                for(int j=1;j<i;j++)
                {
                    if(dcmp(c.o.dis(p[j]),c.r)>0)
                    {
                        c={(p[i]+p[j])/2,p[i].dis(p[j])/2};
                        for(int k=1;k<j;k++)
                        {
                            if(dcmp(c.o.dis(p[k]),c.r)>0)
                            {
                                c=get_circle(p[i],p[j],p[k]);
                            }
                        }
                    }
                }
            }
            w[fir][i]=c.r;
        }
    }
    vector<vector<pair<ld,int>>> ver(n+1);
    for(int last=1;last<=n;last++)
    {
        ver[last].push_back({w[1][last],1});
        for(int i=2;i<=last;i++)
        {
            if(!dcmp(ver[last].back().first,w[i][last])) continue;

            ver[last].push_back({w[i][last],i});
        }
    }
    for(int i=0;i<=n;i++)
        for(int j=0;j<=k;j++)
            dp[i][j]=0x3f3f3f3f3f3f3f3f;
    dp[0][0]=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=min(i,k);j++)
        {
            for(auto [g,id]:ver[i])
            {
                dp[i][j]=min(dp[i][j],dp[id-1][j-1]+g);
            }
        }
    }
    printf("%.10Lf\n",dp[n][k]);
}