Rectangles Too!

A rectangle in the Cartesian plane is given by two corner points $(x_1, y_1)$ and $(x_2, y_2)$, its lower-left and upper-right corners, with $x_1 \le x_2$ and $y_1 \le y_2$.

For two rectangles $A = ((x_1^A, y_1^A), (x_2^A, y_2^A))$ and $B = ((x_1^B, y_1^B), (x_2^B, y_2^B))$, we say that $A$ precedes $B$, written $A \preceq B$, when

$x_2^A < x_1^B \quad\text{and}\quad y_2^A < y_1^B.$

You are given a collection of rectangles in the plane. Find the length $L$ of the longest sequence of rectangles $(A_1, A_2, \ldots, A_L)$ from the collection such that

$A_1 \preceq A_2 \preceq \cdots \preceq A_L.$

The input contains several test cases. Each test case starts with a line containing one integer $n$ ($1 \le n \le 100000$), the number of rectangles. Each of the next $n$ lines contains four integers $x_1\ y_1\ x_2\ y_2$ ($-1000000 \le x_1 \le x_2 \le 1000000$ and $-1000000 \le y_1 \le y_2 \le 1000000$), the lower-left and upper-right corners of one rectangle. The input ends with a line containing a single $0$.

For each test case, print one line with a single integer: the length of the longest chain of rectangles.