Problem: So we got bored and decided to take our own guess at how would "Inception" production go if the budget for the film had been terribly low.

The first scene we remembered was the one that features the whole city bending onto itself:

<image>

It feels like it will require high CGI expenses, doesn't it? Luckily, we came up with a similar-looking scene which was a tiny bit cheaper to make.

Firstly, forget about 3D, that's hard and expensive! The city is now represented as a number line (infinite to make it easier, of course).

Secondly, the city doesn't have to look natural at all. There are n buildings on the line. Each building is a square 1 × 1. Buildings are numbered from 1 to n in ascending order of their positions. Lower corners of building i are at integer points a_i and a_i + 1 of the number line. Also the distance between any two neighbouring buildings i and i + 1 doesn't exceed d (really, this condition is here just to make the city look not that sparse). Distance between some neighbouring buildings i and i + 1 is calculated from the lower right corner of building i to the lower left corner of building i + 1.

Finally, curvature of the bend is also really hard to simulate! Let the bend at some integer coordinate x be performed with the following algorithm. Take the ray from x to +∞ and all the buildings which are on this ray and start turning the ray and the buildings counter-clockwise around point x. At some angle some building will touch either another building or a part of the line. You have to stop bending there (implementing buildings crushing is also not worth its money). 

Let's call the angle between two rays in the final state the terminal angle α_x.

The only thing left is to decide what integer point x is the best to start bending around. Fortunately, we've already chosen m candidates to perform the bending.

So, can you please help us to calculate terminal angle α_x for each bend x from our list of candidates?

Input

The first line contains two integer numbers n and d (1 ≤ n ≤ 2 ⋅ 10^5, 0 ≤ d ≤ 7000) — the number of buildings and the maximum distance between any pair of neighbouring buildings, respectively.

The second line contains n integers a_1, a_2, ..., a_n (a_1 = 0, 0 < a_{i + 1} - a_i ≤ d + 1) — coordinates of left corners of corresponding buildings in ascending order.

The third line contains single integer m (1 ≤ m ≤ 2 ⋅ 10^5) — the number of candidates.

The fourth line contains m integers x_1, x_2, ..., x_m (0 ≤ x_i ≤ a_n + 1, x_i < x_{i + 1}) — the coordinates of bends you need to calculate terminal angles for in ascending order.

Output

Print m numbers. For each bend x_i print terminal angle α_{x_i} (in radians).

Your answer is considered correct if its absolute error does not exceed 10^{-9}.

Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if |a - b| ≤ 10^{-9}.

Examples

Input


3 5
0 5 7
9
0 1 2 3 4 5 6 7 8


Output


1.570796326794897
1.570796326794897
0.785398163397448
0.927295218001612
0.785398163397448
1.570796326794897
1.570796326794897
1.570796326794897
1.570796326794897


Input


2 7
0 4
3
1 3 4


Output


1.570796326794897
0.927295218001612
1.570796326794897


Input


5 0
0 1 2 3 4
6
0 1 2 3 4 5


Output


1.570796326794897
3.141592653589793
3.141592653589793
3.141592653589793
3.141592653589793
1.570796326794897

Note

Here you can see the picture of the city for the first example and the bend at position 2 for it. The angle you need to measure is marked blue. You can see that it's equal to \frac π 4.

You can see that no pair of neighbouring buildings have distance more than 4 between them. d = 4 would also suffice for that test.

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
namespace IO {
const int BUFFER_SIZE = 1 << 15;
char input_buffer[BUFFER_SIZE];
int input_pos = 0, input_len = 0;
char output_buffer[BUFFER_SIZE];
int output_pos = 0;
char number_buffer[100];
uint8_t lookup[100];
void _update_input_buffer() {
  input_len = fread(input_buffer, sizeof(char), BUFFER_SIZE, stdin);
  input_pos = 0;
  if (input_len == 0) input_buffer[0] = EOF;
}
inline char next_char(bool advance = true) {
  if (input_pos >= input_len) _update_input_buffer();
  return input_buffer[advance ? input_pos++ : input_pos];
}
template <typename T>
inline void read_int(T &number) {
  bool negative = false;
  number = 0;
  while (!isdigit(next_char(false)))
    if (next_char() == '-') negative = true;
  do {
    number = 10 * number + (next_char() - '0');
  } while (isdigit(next_char(false)));
  if (negative) number = -number;
}
template <typename T, typename... Args>
inline void read_int(T &number, Args &...args) {
  read_int(number);
  read_int(args...);
}
inline void read_double(double &number) {
  bool negative = false;
  number = 0;
  while (!isdigit(next_char(false)))
    if (next_char() == '-') negative = true;
  do {
    number = 10 * number + (next_char() - '0');
  } while (isdigit(next_char(false)));
  if (next_char(false) == '.') {
    next_char();
    for (double multiplier = 0.1; isdigit(next_char(false)); multiplier *= 0.1)
      number += multiplier * (next_char() - '0');
  }
  if (negative) number = -number;
}
void _flush_output() {
  fwrite(output_buffer, sizeof(char), output_pos, stdout);
  output_pos = 0;
}
inline void write_char(char c) {
  if (output_pos == BUFFER_SIZE) _flush_output();
  output_buffer[output_pos++] = c;
}
template <typename T>
inline void write_int(T number, char after = '\0') {
  if (number < 0) {
    write_char('-');
    number = -number;
  }
  int length = 0;
  while (number >= 10) {
    uint8_t lookup_value = lookup[number % 100];
    number /= 100;
    number_buffer[length++] = (lookup_value & 15) + '0';
    number_buffer[length++] = (lookup_value >> 4) + '0';
  }
  if (number != 0 || length == 0) write_char(number + '0');
  for (int i = length - 1; i >= 0; i--) write_char(number_buffer[i]);
  if (after) write_char(after);
}
inline void write_double(double number, char after = '\0', int places = 6) {
  if (number < 0) {
    write_char('-');
    number = -number;
  }
  assert(number <= 9e18);
  number += 0.5 * pow(0.1, places);
  long long floored = (long long)number;
  write_int(floored);
  if (places > 0) {
    write_char('.');
    number = (number - floored) * pow(10.0, places);
    floored = (long long)number;
    int length = 0;
    for (int i = 0; i < places / 2; i++) {
      uint8_t lookup_value = lookup[floored % 100];
      floored /= 100;
      number_buffer[length++] = (lookup_value & 15) + '0';
      number_buffer[length++] = (lookup_value >> 4) + '0';
    }
    if (places % 2) number_buffer[length++] = floored + '0';
    assert(length == places);
    for (int i = places - 1; i >= 0; i--) write_char(number_buffer[i]);
  }
  if (after) write_char(after);
}
void init() {
  bool exit_success = atexit(_flush_output) == 0;
  assert(exit_success);
  for (int i = 0; i < 100; i++) lookup[i] = (i / 10 << 4) + i % 10;
}
}  // namespace IO
const int OFFSET = 5;
const int D_MAX = 7050;
const int INF = 1e9 + 5;
int N;
vector<int> A;
double angle_save[D_MAX];
inline double single_angle(int distance) {
  if (angle_save[distance] < 0) angle_save[distance] = atan2(1, distance);
  return angle_save[distance];
}
double solve(int x) {
  static int index = 0;
  while (index < N && A[index] < x) index++;
  int left = index - 1, right = index;
  int left_dist = INF, right_dist = INF;
  double answer = 0;
  if (left >= 0) left_dist = x - 1 - A[left];
  if (right < N) right_dist = A[right] - x;
  int start_dist = min(left_dist, right_dist);
  answer = single_angle(start_dist);
  while (left >= 0 && right < N) {
    left_dist = x - 1 - A[left];
    right_dist = A[right] - x;
    if (max(left_dist, right_dist) > 2 * start_dist + OFFSET) break;
    if (abs(left_dist - right_dist) <= 1) {
      answer = max(answer, 2 * single_angle(max(left_dist, right_dist)));
      break;
    }
    if (left_dist < right_dist)
      left--;
    else
      right++;
  }
  return answer;
}
int main() {
  IO::init();
  fill(angle_save, angle_save + D_MAX, -1);
  int D;
  IO::read_int(N, D);
  A.resize(N);
  for (int &a : A) IO::read_int(a);
  int M;
  IO::read_int(M);
  while (M-- > 0) {
    int x;
    IO::read_int(x);
    IO::write_double(solve(x), '\n', 9);
  }
}