
import eulerlib


# Suppose the box has b blue discs and r red discs.
# The probability of taking 2 blue discs is [b / (b + r)] * [(b - 1) / (b + r - 1)],
# which we want to be equal to 1/2. Rearrange the equation:
#   [b(b - 1)] / [(b + r)(b + r - 1)] = 1 / 2.
#   2b(b - 1) = (b + r)(b + r - 1).
#   2b^2 - 2b = b^2 + br - b + br + r^2 - r.
#   b^2 - b = r^2 + 2br - r.
#   b^2 - (2r + 1)b + (r - r^2) = 0.
# Apply the quadratic equation to solve for b:
#   b = [(2r + 1) +/- sqrt((2r + 1)^2 - 4(r - r^2))] / 2
#     = r + [1 +/- sqrt(8r^2 + 1)]/2
#     = r + [sqrt(8r^2 + 1) + 1]/2.  (Discard the minus solution because it would make b < r)
# 
# For b to be an integer, we need sqrt(8r^2 + 1) to be odd, and also 8r^2 + 1 be a perfect square.
# Assume 8y^2 + 1 = x^2 for some integer x > 0.
# We can see this is in fact a Pell's equation: x^2 - 8y^2 = 1.
# 
# Suppose we have the solution (x0, y0) such that x0 > 0 and x0 is as small as possible.
# This is called the fundamental solution, and all other solutions be derived from it (proven elsewhere).
# Suppose (x0, y0) and (x1, y1) are solutions. Then we have:
#   x0^2 - 8*y0^2 = 1.
#   (x0 + y0*sqrt(8))(x0 - y0*sqrt(8)) = 1.
#   (x1 + y1*sqrt(8))(x1 - y1*sqrt(8)) = 1.  (Similarly)
# Multiply them together:
#   [(x0 + y0*sqrt(8))(x0 - y0*sqrt(8))][(x1 + y1*sqrt(8))(x1 - y1*sqrt(8))] = 1 * 1.
#   [(x0 + y0*sqrt(8))(x1 + y1*sqrt(8))][(x0 - y0*sqrt(8))(x1 - y1*sqrt(8))] = 1.
#   [x0*x1 + x0*y1*sqrt(8) + x1*y0*sqrt(8) + 8y0*y1][x0*x1 - x0*y1*sqrt(8) - x1*y0*sqrt(8) + 8y0*y1] = 1.
#   [(x0*x1 + 8y0*y1) + (x0*y1 + x1*y0)*sqrt(8)][(x0*x1 + 8y0*y1) - (x0*y1 + x1*y0)*sqrt(8)] = 1.
#   (x0*x1 + 8y0*y1)^2 - 8*(x0*y1 + x1*y0)^2 = 1.
# Therefore (x0*x1 + 8y0*y1, x0*y1 + x1*y0) is also a solution.
# By inspection, the fundamental solution is (3, 1).
def compute():
	# Fundamental solution
	x0 = 3
	y0 = 1
	
	# Current solution
	x = x0
	y = y0  # An alias for the number of red discs
	while True:
		# Check if this solution is acceptable
		sqrt = eulerlib.sqrt(y**2 * 8 + 1)
		if sqrt % 2 == 1:  # Is odd
			blue = (sqrt + 1) // 2 + y
			if blue + y > 10**12:
				return str(blue)
		
		# Create the next bigger solution
		nextx = x * x0 + y * y0 * 8
		nexty = x * y0 + y * x0
		x = nextx
		y = nexty


if __name__ == "__main__":
	print(compute())
