Problem: In some country live wizards. They love to ride trolleybuses.

A city in this country has a trolleybus depot with n trolleybuses. Every day the trolleybuses leave the depot, one by one and go to the final station. The final station is at a distance of d meters from the depot. We know for the i-th trolleybus that it leaves at the moment of time ti seconds, can go at a speed of no greater than vi meters per second, and accelerate with an acceleration no greater than a meters per second squared. A trolleybus can decelerate as quickly as you want (magic!). It can change its acceleration as fast as you want, as well. Note that the maximum acceleration is the same for all trolleys.

Despite the magic the trolleys are still powered by an electric circuit and cannot overtake each other (the wires are to blame, of course). If a trolleybus catches up with another one, they go together one right after the other until they arrive at the final station. Also, the drivers are driving so as to arrive at the final station as quickly as possible.

You, as head of the trolleybuses' fans' club, are to determine for each trolley the minimum time by which it can reach the final station. At the time of arrival at the destination station the trolleybus does not necessarily have zero speed. When a trolley is leaving the depot, its speed is considered equal to zero. From the point of view of physics, the trolleybuses can be considered as material points, and also we should ignore the impact on the speed of a trolley bus by everything, except for the acceleration and deceleration provided by the engine.

Input

The first input line contains three space-separated integers n, a, d (1 ≤ n ≤ 105, 1 ≤ a, d ≤ 106) — the number of trolleybuses, their maximum acceleration and the distance from the depot to the final station, correspondingly.

Next n lines contain pairs of integers ti vi (0 ≤ t1 < t2... < tn - 1 < tn ≤ 106, 1 ≤ vi ≤ 106) — the time when the i-th trolleybus leaves the depot and its maximum speed, correspondingly. The numbers in the lines are separated by spaces.

Output

For each trolleybus print a single line the time it arrives to the final station. Print the times for the trolleybuses in the order in which the trolleybuses are given in the input. The answer will be accepted if the absolute or relative error doesn't exceed 10 - 4.

Examples

Input

3 10 10000
0 10
5 11
1000 1


Output

1000.5000000000
1000.5000000000
11000.0500000000


Input

1 2 26
28 29


Output

33.0990195136

Note

In the first sample the second trolleybus will catch up with the first one, that will happen at distance 510.5 meters from the depot. The trolleybuses will go the remaining 9489.5 meters together at speed 10 meters per second. As a result, both trolleybuses will arrive to the final station by the moment of time 1000.5 seconds. The third trolleybus will not catch up with them. It will arrive to the final station by the moment of time 11000.05 seconds.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<ld, ld>;
template <class T>
using vec = vector<T>;
using vb = vec<bool>;
using vi = vec<int>;
using vl = vec<ll>;
using vd = vec<ld>;
using vs = vec<str>;
using vpi = vec<pi>;
using vpl = vec<pl>;
using vpd = vec<pd>;
template <class Key, class T, class Hash = hash<Key>,
          class Pred = equal_to<Key>,
          class Alloc = allocator<pair<const Key, T>>>
using umap = unordered_map<Key, T, Hash, Pred, Alloc>;
template <class Key, class Hash = hash<Key>, class Pred = equal_to<Key>,
          class Alloc = allocator<Key>>
using uset = unordered_set<Key, Hash, Pred, Alloc>;
int MOD = 1e9 + 7, MX = 5e5 + 5;
ll INF = 1e18;
ld PI = acos((ld)-1), EPS = 1e-8;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
namespace input {
template <class T>
inline void re(complex<T>& first);
template <class T1, class T2>
inline void re(pair<T1, T2>& p);
template <class T>
inline void re(vector<T>& a);
template <class T, size_t SZ>
inline void re(array<T, SZ>& a);
template <class T>
inline void re(T& first) {
  cin >> first;
}
inline void re(double& first) {
  string t;
  re(t);
  first = stod(t);
}
inline void re(ld& first) {
  string t;
  re(t);
  first = stold(t);
}
template <class T, class... Ts>
inline void re(T& t, Ts&... ts) {
  re(t);
  re(ts...);
}
template <class T>
inline void re(complex<T>& first) {
  T a, b;
  re(a, b);
  first = {a, b};
}
template <class T1, class T2>
inline void re(pair<T1, T2>& p) {
  re(p.f, p.s);
}
template <class T>
inline void re(vector<T>& a) {
  for (int i = (0); i != ((int)(a.size())); i += (0 < (int)(a.size()) ? 1 : -1))
    re(a[i]);
}
template <class T, size_t SZ>
inline void re(array<T, SZ>& a) {
  for (int i = (0); i != (SZ); i += (0 < SZ ? 1 : -1)) re(a[i]);
}
inline void setin(string s) { freopen(s.c_str(), "r", stdin); }
}  // namespace input
namespace output {
inline void pr(int first) { cout << first; }
inline void pr(long first) { cout << first; }
inline void pr(ll first) { cout << first; }
inline void pr(unsigned first) { cout << first; }
inline void pr(unsigned long first) { cout << first; }
inline void pr(unsigned long long first) { cout << first; }
inline void pr(float first) { cout << first; }
inline void pr(double first) { cout << first; }
inline void pr(ld first) { cout << first; }
inline void pr(char first) { cout << first; }
inline void pr(const char* first) { cout << first; }
inline void pr(const string& first) { cout << first; }
inline void pr(bool first) { pr(first ? "true" : "false"); }
template <class T>
inline void pr(const complex<T>& first) {
  cout << first;
}
template <class T1, class T2>
inline void pr(const pair<T1, T2>& first);
template <class T>
inline void pr(const T& first);
template <class T, class... Ts>
inline void pr(const T& t, const Ts&... ts) {
  pr(t);
  pr(ts...);
}
template <class T1, class T2>
inline void pr(const pair<T1, T2>& first) {
  pr("{", first.first, ", ", first.second, "}");
}
template <class T>
inline void pr(const T& first) {
  pr("{");
  bool fst = 1;
  for (const auto& a : first) pr(!fst ? ", " : "", a), fst = 0;
  pr("}");
}
inline void ps() { pr("\n"); }
template <class T, class... Ts>
inline void ps(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) pr(" ");
  ps(ts...);
}
inline void pc() { pr("]\n"); }
template <class T, class... Ts>
inline void pc(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) pr(", ");
  pc(ts...);
}
inline void setout(string s) { freopen(s.c_str(), "w", stdout); }
}  // namespace output
using namespace input;
using namespace output;
void solve();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  solve();
  return 0;
}
int N, A, D, T, V;
void solve() {
  re(N, A, D);
  ld lf = 0;
  for (int i = (0); i != (N); i += (0 < N ? 1 : -1)) {
    re(T, V);
    ld at = (ld)V / A;
    ld d = .5 * A * at * at;
    0;
    ld t;
    if (d <= D)
      t = (D - d) / V + at;
    else
      t = sqrt(2.0 * D / A);
    ps(max(lf, T + t));
    ckmax(lf, T + t);
  }
}