#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
int N,L;
vpi v;
vector<ld> tmp;
ld perp(pd x, pd y) {
return (y.f*y.f+y.s*y.s-x.f*x.f-x.s*x.s)/(2*y.f-2*x.f);
}
ld dist(pd a, pd b) {
return sqrt(pow(a.f-b.f,2)+pow(a.s-b.s,2));
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N >> L;
F0R(i,N) {
int x,y; cin >> x >> y; y = abs(y);
if (sz(v) && v.back().f == x) {
if (y >= v.back().s) continue;
else {
v.pop_back();
if (sz(tmp)) tmp.pop_back();
}
}
while (sz(tmp) && tmp.back() >= perp(v.back(),{x,y})) {
v.pop_back();
tmp.pop_back();
}
if (sz(v)) tmp.pb(perp(v.back(),{x,y}));
v.pb({x,y});
}
ld ans = 0;
F0R(i,sz(v)) {
ld mn = (i == 0 ? -INF : tmp[i-1]);
ld mx = (i == sz(v)-1 ? INF : tmp[i]);
if (mx < 0 || mn > L) continue;
mn = max(mn,(ld)0), mx = min(mx,(ld)L);
ans = max(ans,dist(v[i],{mn,0}));
ans = max(ans,dist(v[i],{mx,0}));
}
cout << fixed << setprecision(6) << ans;
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
