Problem:
Score: 300 points

Problem Statement

Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.

An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.

At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?

Constraints

All values in input are integers.

1 \leq A, B \leq 1000

0 \leq H \leq 11

0 \leq M \leq 59

Input

Input is given from Standard Input in the following format:

A B H M

Output

Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.

Sample Input 1

3 4 9 0

Sample Output 1

5.00000000000000000000

The two hands will be in the positions shown in the figure below, so the answer is 5 centimeters.

Sample Input 2

3 4 10 40

Sample Output 2

4.56425719433005567605

The two hands will be in the positions shown in the figure below. Note that each hand always rotates at constant angular velocity.

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define PI 3.141592

int main() {
  short A, B, H, M;
  scanf("%hd%hd%hd%hd", &A, &B, &H, &M);

  // 所与の時間より基準分を求める
  short base_minutes = H * 5; // 1時だったら5分が基準分

  // 差分の分数を求める
  short diff_minutes = abs(M - base_minutes); // 1分が所与だったら、4分（絶対値）、7分が所与だったら2分
  if (diff_minutes - 30 >= 0) {
    diff_minutes = diff_minutes - 30;
  }

  /* printf("%s", "diff_minutes:"); */
  /* printf("%hd", diff_minutes); */
  /* printf("%c", '\n'); */

  // 差分の分から角度を求める
  short degree = diff_minutes * 6; // 1分辺りの角度が6度なので、差が15分だったら15*6で90度となる

  /* printf("%s", "degree:"); */
  /* printf("%hd", degree); */
  /* printf("%c", '\n'); */

  // 角度からラジアンを求める
  double radian = degree * PI / 180.0;

  /* printf("%s", "A*A:"); */
  /* printf("%d", A*A); */
  /* printf("%c", '\n'); */

  /* printf("%s", "B*B:"); */
  /* printf("%d", B*B); */
  /* printf("%c", '\n'); */

  /* printf("%s", "2*A*B:"); */
  /* printf("%d", 2*A*B); */
  /* printf("%c", '\n'); */

  /* printf("%s", "cos(radian):"); */
  /* printf("%lf", cos(radian)); */
  /* printf("%c", '\n'); */

  /* printf("%s", "((2 * A * B) * cos(radian)):"); */
  /* printf("%lf", ((2 * A * B) * cos(radian))); */
  /* printf("%c", '\n'); */

  // 余弦定理でAとBの長さから問われている辺の長さを求める
  int x = ((2 * A * B) * cos(radian));
  double output = (A * A) + (B * B) - x;

  printf("%lf", output);
  printf("%c", '\n');

  return 0;
}