#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mp make_pair
#define pb push_back
#define x first
#define y second
#define pii pair<int, int>
#define p3i pair<pii, int>
#define pll pair<ll, ll>
#define p3l pair<pll, ll>
#define vi vector<int>
#define vpii vector<pii>
#define vp3i vector<p3i>
#define vpll vector<pll>
#define vp3l vector<p3l>
#define lseg L, (L+R)/2, N*2+1
#define rseg (L+R)/2+1, R, N*2+2
#define ub upper_bound
#define lb lower_bound
#define pq priority_queue
#define MN 1000000007
#define fox(k, x) for (int k=0; k<x; ++k)
#define fox1(k, x) for (int k=1; k<=x; ++k)
#define foxr(k, x) for (int k=x-1; k>=0; --k)
#define fox1r(k, x) for (int k=x; k>0; --k)
#define ms multiset
#define flood(x) memset(x, 0x3f3f3f3f, sizeof x)
#define drain(x) memset(x, 0, sizeof x)
#define rng() ((rand() << 14)+rand())
#define scan(X) do{while((X=getchar())<'0'); for(X-='0'; '0'<=(_=getchar()); X=(X<<3)+(X<<1)+_-'0');}while(0)
char _;
#define pi 3.14159265358979323846
#define startrng mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long double ans;
int n, d, x, y;
vector<pll> v;
vector<pll> hull;
//f(x) = x^2 + mx + b
//xa1+b1=xa2+b2
//x=(b2-b1)/(a1-a2);
long double eval(pll A, long double X){
//cout << A.x << ' ' << A.y << ' ' << X << endl;
return sqrt(X*X+X*A.x+A.y);
}
void check(pll A, pll B){
long double x=(B.y-A.y)/((long double)(A.x-B.x));
if (x<0 || x>d) return;
ans = max(eval(A, x), ans);
}
bool cross(pll A, pll O, pll B){
return ((long double)(A.x-O.x))*(B.y-O.y)>=((long double)(A.y-O.y))*(B.x-O.x);//+(1e-9);
}
int32_t main(){
scanf("%i%i", &n, &d);
fox(l, n){
scanf("%i%i", &x, &y);
v.pb(mp(-2*x, 1LL*y*y+1LL*x*x));
}
sort(v.begin(), v.end());
for (auto i:v){
while(hull.size()>1 && cross(hull[hull.size()-2], hull.back(), i)){
hull.pop_back();
}
hull.pb(i);
}
fox(l, hull.size()-1){
check(hull[l], hull[l+1]);
}
//cout << hull.size() << endl;
//cout << ans << endl;
long double lft=(1LL<<62), rit=(1LL<<62);
fox(l, hull.size()){
lft=min(lft, eval(hull[l], 0));
rit=min(rit, eval(hull[l], d));
}
//cout << lft << ' ' << rit << endl;
ans = max(ans, max(lft, rit));
//cout << sqrt(ans) << endl;
printf("%.8Lf", ans);
return 0;
}
/*
3 10
0 0
5 1000000000
11 1
*/
