The Arc That Stops the Circle

A square of side S and a circle of radius S / 2 are rolling along the same road. The road is not flat: it carries two arcs joined end to end, rising out of the ground like two low humps side by side.

When the circle reaches the point where the two arcs meet, the elevation in front of it is too steep to climb, so the circle stops advancing and simply spins in place. The square, however, keeps rolling over the arcs without trouble.

The arcs are shaped so that this is exactly what happens, and so that the square fits them perfectly: as the square tips across an arc, its two lower corners come to rest precisely on the two ends of that arc.

Given S, report the height of the top of the arc above the road — the distance from the highest point of the arc down to the flat road it rises from.

Input Format

The first line of input contains a single integer S, the side of the square.

Output Format

Print a single number, the height of the top of the arc above the road.

Print it with exactly 4 digits after the decimal point, always — including trailing zeros. A value of 2 must be printed as 2.0000, and a value of 0.41421356... must be printed as 0.4142. Do not print more than 4 decimals and do not drop a trailing 0.

Rounding is round-half-to-even at the 4th decimal place. (inferred — the source only says "up to 4 decimal places"; the choice is not observable here, because within the stated constraints the exact answer is irrational and never lands on a half-way point, so round-half-up and round-half-to-even always agree)

Constraints

1 <= S <= 100000 (inferred — the source states no bounds. S is the side of a square, so it is a positive integer; 0 and negative values are not valid input. The upper bound is chosen so that the answer stays far from any 4th-decimal rounding boundary: over the whole range the closest any exact answer comes to a half-way point is about 473 units in the last place of a double, so every correct double implementation prints the same digits.)
S is an integer.

Examples

Example 1

Input:

Output:

1.6569

Explanation: The square has side 8. Laid across the arc, its two lower corners land on the two ends of the arc, which are 8 apart, while the square's own diagonal measures 11.3137. The diagonal overhangs the arc by 11.3137 - 8 = 3.3137, split evenly between the two ends, and the leftover at one end is the clearance the arc must rise to: 1.6569.

Example 2

Input:

Output:

0.4142

Explanation: The same picture at side 2. The arc's ends are 2 apart and the square's diagonal is 2.8284, so the overhang is 2.8284 - 2 = 0.8284. Halving it between the two ends leaves 0.4142 at each end, which is the height of the arc.

Notes

The answer does not depend on the circle's radius. The circle only fixes which arc is being described — the arc is exactly steep enough to stall a circle of radius S / 2 — and the height itself is then determined entirely by how the square of side S sits on that arc.
The answer grows linearly with S: doubling the side doubles the height. Example 1 (S = 8) is four times Example 2 (S = 2).
The output is always strictly positive, and it is always smaller than S. The smallest legal input S = 1 still gives a positive height (0.2071), so there is no degenerate case where the arc flattens to nothing. (inferred — follows from S >= 1)
Formatting is part of the answer. Print exactly 4 decimals with a plain fixed-point format (printf("%.4f", ...) in C/C++, String.format(Locale.US, "%.4f", ...) in Java, "%.4f" % x in Python). Scientific notation, 4 significant digits, or a language default float repr will all be judged wrong.
Use double-precision floating point. Single precision is not accurate enough at the top of the range.
Trailing whitespace and the trailing newline are not significant.