Absolute-Tolerance Comparison

Version: Phuoc Hung OJ Extended

Problem Statement

Given real numbers $a$, $b$ and a nonnegative absolute tolerance $eps$, regard the two numbers as close when $|a-b|\le eps$. Report the result of Python 3 math.isclose(a, b, rel_tol=0.0, abs_tol=eps); no relative tolerance is used.

Input

The only line contains real numbers $a$, $b$ and absolute tolerance $eps$, in that order.

Output

Print exactly True if $|a-b|\le eps$, otherwise False.

Subtasks

• Subtask 1 (100 points): $a,b,eps$ are finite and $eps\ge0$.

Examples

Example 1

Input

0.30000000000000004 0.3 1e-09

Output

True

Explanation. The computed difference is approximately 5.55111512313e-17 and the permitted tolerance is 1e-09. The difference is ≤ the tolerance, so the answer is True.

Example 2

Input

1.0 1.0000000001 1e-08

Output

True

Explanation. The computed difference is approximately 1.00000008274e-10 and the permitted tolerance is 1e-08. The difference is ≤ the tolerance, so the answer is True.

Example 3

Input

1000 1000.01 0.0001

Output

False

Explanation. The computed difference is approximately 0.00999999999999 and the permitted tolerance is 0.0001. The difference is > the tolerance, so the answer is False.