Relative-Tolerance Comparison

Version: Phuoc Hung OJ Extended

Problem Statement

Given real numbers $a$, $b$ and a relative tolerance $eps$, regard them as close when $|a-b|\le eps\max(|a|,|b|)$. Use the semantics of Python 3 math.isclose(a, b, rel_tol=eps, abs_tol=0.0). There is no additional absolute tolerance near zero.

Input

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

Output

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

Subtasks

• Subtask 1 (100 points): $a,b,eps$ are finite; $0\le eps<1$.

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 3e-10. The difference is ≤ the tolerance, so the answer is True.

Example 2

Input

1 1.0000000001 1e-08

Output

True

Explanation. The computed difference is approximately 1.00000008274e-10 and the permitted tolerance is 1.0000000001e-08. The difference is ≤ the tolerance, so the answer is True.

Example 3

Input

1000 1000.01 0.0001

Output

True

Explanation. The computed difference is approximately 0.00999999999999 and the permitted tolerance is 0.100001. The difference is ≤ the tolerance, so the answer is True.