Stern-Brocot tree

The Stern-Brocot tree is an infinite complete binary tree in which the vertices correspond one-for-one to the positive rational numbers, whose values are ordered from the left to the right as in a search tree.



Figure 1 shows a part of the Stern-Brocot tree, which has the first 4 rows. Each node in the tree is marked in a red cycle. The value in the node is the mediant of the left and right fractions. The mediant of two fractions A/B and C/D is defined as (A+C)/(B+D).
To construct the Stern-Brocot tree, we first define the left fraction of the root node is 0/1, and the right fraction of the root node is 1/0. So the value in the root node is the mediant of 0/1 and 1/0, which is (0+1)/(1+0)=1/1. Then the value of root node becomes the right fraction of the left child, and the left fraction of the right child. For example, the 1st node in row2 has 0/1 as its left fraction and 1/1(which is the value of its parent node) as its right fraction. So the value of the 1st node in row2 is (0+1)/(1+1)=1/2. For the same reason, the value of the 2nd node in row2 is (1+1)/(1+0)=2/1. This construction progress goes on infinitly. As a result, every positive rational number can be found on the Stern-Brocot tree, and can be found only once.

Given a rational number in form of P/Q, find the position of P/Q in the Stern-Brocot Tree.

Input consists of two integers, P and Q (1<=P,Q<=1000), which represent the rational number P/Q. We promise P and Q are relatively prime.

Output consists of two integers, R and C.
R indicates the row index of P/Q in the Stern-Brocot Tree, C indicates the index of P/Q in the row.
Both R and C are base 1.
We promise the position of P/Q is always in the first 12 rows of the Stern-Brocot tree, which means R<=12.

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