There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

<image>

You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 
#include <bits/stdc++.h>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("sse4")
template <typename T>
inline T max(T a, T b) {
  return a > b ? a : b;
}
template <typename T>
inline T min(T a, T b) {
  return a < b ? a : b;
}
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline bool repr(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool repl(T &a, T b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline T gcd(T a, T b) {
  T t;
  if (a < b) {
    while (a) {
      t = a;
      a = b % a;
      b = t;
    }
    return b;
  } else {
    while (b) {
      t = b;
      b = a % b;
      a = t;
    }
    return a;
  }
}
template <typename T>
inline T sqr(T x) {
  return x * x;
}
const int N = 2e5 + 7;
double ht, a, b;
int n, q;
double left[N], right[N];
double sum[N];
double res[N];
struct {
  double x, y;
} ps[N];
inline double project(const std::pair<int, int> po, int s) {
  return double(po.first - s) * ht / po.second + s;
}
inline std::pair<double, double> project(const std::pair<int, int> po,
                                         const std::pair<int, int> fe) {
  return std::pair<double, double>(project(po, fe.first),
                                   project(po, fe.second));
}
inline double rproject(const std::pair<double, double> po, double a) {
  return -ht * (po.first - a) / (po.second - ht) + a;
}
inline std::pair<double, double> rproject(const std::pair<double, double> po) {
  return std::pair<double, double>(rproject(po, a), rproject(po, b));
}
inline double overlap(const std::pair<double, double> range) {
  return max(0., min(b + 0., range.second) - max(a + 0., range.first));
}
template <class T = int>
inline T readInt();
inline double readDouble();
inline int readUInt();
inline int readChar();
inline void readWord(char *s);
inline bool readLine(char *s);
inline bool isEof();
inline int getChar();
inline int peekChar();
inline bool seekEof();
inline void skipBlanks();
template <class T>
inline void writeInt(T x, char end = 0, int len = -1);
inline void writeChar(int x);
inline void writeWord(const char *s);
inline void writeDouble(double x, int len = 0);
inline void flush();
static const int buf_size = 4096;
static unsigned char buf[buf_size];
static int buf_len = 0, buf_pos = 0;
inline bool isEof() {
  if (buf_pos == buf_len) {
    buf_pos = 0, buf_len = fread(buf, 1, buf_size, stdin);
    if (buf_pos == buf_len) return 1;
  }
  return 0;
}
inline int getChar() { return isEof() ? -1 : buf[buf_pos++]; }
inline int peekChar() { return isEof() ? -1 : buf[buf_pos]; }
inline bool seekEof() {
  int c;
  while ((c = peekChar()) != -1 && c <= 32) buf_pos++;
  return c == -1;
}
inline void skipBlanks() {
  while (!isEof() && buf[buf_pos] <= 32U) buf_pos++;
}
inline int readChar() {
  int c = getChar();
  while (c != -1 && c <= 32) c = getChar();
  return c;
}
inline int readUInt() {
  int c = readChar(), x = 0;
  while ('0' <= c && c <= '9') x = x * 10 + c - '0', c = getChar();
  return x;
}
template <class T>
inline T readInt() {
  int s = 1, c = readChar();
  T x = 0;
  if (c == '-')
    s = -1, c = getChar();
  else if (c == '+')
    c = getChar();
  while ('0' <= c && c <= '9') x = x * 10 + c - '0', c = getChar();
  return s == 1 ? x : -x;
}
inline double readDouble() {
  int s = 1, c = readChar();
  double x = 0;
  if (c == '-') s = -1, c = getChar();
  while ('0' <= c && c <= '9') x = x * 10 + c - '0', c = getChar();
  if (c == '.') {
    c = getChar();
    double coef = 1;
    while ('0' <= c && c <= '9') x += (c - '0') * (coef *= 1e-1), c = getChar();
  }
  return s == 1 ? x : -x;
}
inline void readWord(char *s) {
  int c = readChar();
  while (c > 32) *s++ = c, c = getChar();
  *s = 0;
}
inline bool readLine(char *s) {
  int c = getChar();
  while (c != '\n' && c != -1) *s++ = c, c = getChar();
  *s = 0;
  return c != -1;
}
static int write_buf_pos = 0;
static char write_buf[buf_size];
static struct buffer_flusher_t {
  ~buffer_flusher_t() { flush(); }
} buffer_flusher;
inline void writeChar(int x) {
  if (write_buf_pos == buf_size)
    fwrite(write_buf, 1, buf_size, stdout), write_buf_pos = 0;
  write_buf[write_buf_pos++] = x;
}
inline void flush() {
  if (write_buf_pos) {
    fwrite(write_buf, 1, write_buf_pos, stdout), write_buf_pos = 0;
    fflush(stdout);
  }
}
template <class T>
inline void writeInt(T x, char end, int output_len) {
  if (x < 0) writeChar('-'), x = -x;
  char s[24];
  int n = 0;
  while (x || !n) s[n++] = '0' + x % 10, x /= 10;
  while (n < output_len) s[n++] = '0';
  while (n--) writeChar(s[n]);
  if (end) writeChar(end);
}
inline void writeWord(const char *s) {
  while (*s) writeChar(*s++);
}
inline void writeDouble(double x, int output_len) {
  if (x < 0) writeChar('-'), x = -x;
  int t = (long long)x;
  writeInt<long long>(t), x -= t;
  writeChar('.');
  for (int i = output_len - 1; i > 0; i--) {
    x *= 10;
    t = std::min(9, (int)x);
    writeChar('0' + t), x -= t;
  }
  x *= 10;
  t = std::min(9, (int)(x + 0.5));
  writeChar('0' + t);
}
int main() {
  double x, y;
  ht = readInt();
  a = readInt();
  b = readInt();
  n = readInt();
  for (int i = 1, iend = n; i <= iend; i++) {
    left[i] = readInt();
    right[i] = readInt();
  }
  for (int i = n; i; i--) {
    sum[i] = sum[i + 1] + (right[i] - left[i]);
  }
  left[0] = right[0] = -2e9;
  left[n + 1] = right[n + 1] = 2e9;
  q = readInt();
  for (int i = 1, iend = q; i <= iend; i++) {
    x = readInt();
    y = readInt();
    ps[i].x = x;
    ps[i].y = y;
  }
  for (int i = 1, iend = q; i <= iend; i++) {
    double res = 0;
    x = ps[i].x, y = ps[i].y;
    double first, second;
    first = ht * (a - x) / (y - ht) + a;
    second = ht * (b - x) / (y - ht) + b;
    if (first >= right[n] || second <= left[1]) {
    }
    int l = n + 1;
    int tx, ty;
    for (tx = 1; tx < l;) {
      ty = (tx + l) >> 1;
      if (first <= right[ty])
        l = ty;
      else
        tx = ty + 1;
    }
    int r = n + 1;
    for (tx = l - 1; tx < r;) {
      ty = (tx + r) >> 1;
      if (second < left[ty])
        r = ty;
      else
        tx = ty + 1;
    }
    --r;
    if (r < l) {
    } else {
      res = sum[l] - sum[r + 1];
      if (first > left[l]) res -= (first - left[l]);
      if (right[r] > second) res -= (right[r] - second);
      res *= (y - ht) / y;
    }
    ps[i].x = res;
  }
  for (int i = 1, iend = q; i <= iend; i++) {
    writeDouble(ps[i].x, 15);
    writeChar('\n');
  }
}
