#MTH000000021. Trở lại toán trung cấp (Back to Intermediate Math)

Trở lại toán trung cấp (Back to Intermediate Math)

Back to Intermediate Math

Source: UVa

Version: Phuoc Hung OJ Extended

Problem Statement

You want to cross a river of width dd metres. The current flows parallel to the bank at speed vv m/s, while the boat's speed relative to the water is uu m/s.

There are two objectives:

  • Fastest crossing: point the boat perpendicular to the bank so the cross-river component is maximal; the current may carry the boat downstream.
  • Shortest path: point the boat upstream just enough that the actual trajectory is perpendicular to the bank.

If these two different crossings are both well-defined, compute the shortest-path time minus the fastest-crossing time. Otherwise report that the value cannot be determined.

Input

One line contains three nonnegative real numbers dd, vv, uu, with d>0d>0.

Output

If two different crossings as described cannot be determined, print

can't determine

Otherwise print the time difference with exactly three digits after the decimal point.

Subtasks

  • Subtask 1 (20%): Only cases where two distinct crossings cannot be determined.
  • Subtask 2 (30%): d,v,ud,v,u are integers and two distinct crossings exist.
  • Subtask 3 (50%): Full source domain: d>0d>0, v≥0v\ge0, u≥0u\ge0.

Examples

Input

8 5 6

Output

1.079

Explanation

For d=8d=8, v=5v=5, and u=6u=6, the fastest crossing time is

tf=86≈1.333333.t_f=\frac{8}{6}\approx1.333333.

For the path whose actual trajectory is perpendicular to the bank, the cross-river speed is

62−52=11.\sqrt{6^2-5^2}=\sqrt{11}.

Hence that crossing time is

ts=811≈2.412091.t_s=\frac{8}{\sqrt{11}}\approx2.412091.

The difference is ts−tf≈1.078758t_s-t_f\approx1.078758, which rounds to 1.079.