Problem: 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}.

Examples

Input

3 4 9 0


Output

5.00000000000000000000


Input

3 4 10 40


Output

4.56425719433005567605

Solution (in PYTHON):

Solution:
FAST_IO = 0
if FAST_IO:
    import io, sys, atexit
    rr = iter(sys.stdin.read().splitlines()).next
    sys.stdout = _OUTPUT_BUFFER = io.BytesIO()
    @atexit.register
    def write():
        sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
else:
    rr = raw_input
rri = lambda: int(rr())
rrm = lambda: map(int, rr().split())

####
import math
def solve(A,B,H,M):
    # A = hour length, B = min length
    t = 60 * H + M
    # First point is at t % 12*60) / 12*60
    # Second point is at t % 60 ) / 60
    #print("!", t)
    alpha = t % (12 * 60)
    alpha /= (12.0 * 60.0)
    alpha = (0.25 - alpha) % 1.00
    #print("!", alpha)
    alpha *= math.pi * 2

    beta = t % (60)
    beta /= (60.0)
    beta = (0.25 - beta) % 1.00
    #print("!", beta)
    beta *= math.pi * 2

    x1, y1 = A * math.cos(alpha), A * math.sin(alpha)
    x2, y2 = B * math.cos(beta), B * math.sin(beta)
    d = (x2-x1)**2 + (y2-y1)**2
    d = d ** 0.5
    return d

print solve(*rrm())