Description

Background 
Binary trees are a common data structure in computer science. In this problem we will look at an infinite binary tree where the nodes contain a pair of integers. The tree is constructed like this: 

• The root contains the pair (1, 1). 
• If a node contains (a, b) then its left child contains (a + b, b) and its right child (a, a + b)

Problem 
Given the contents (a, b) of some node of the binary tree described above, suppose you are walking from the root of the tree to the given node along the shortest possible path. Can you find out how often you have to go to a left child and how often to a right child?

Input

The first line contains the number of scenarios. 
Every scenario consists of a single line containing two integers i and j (1 <= i, j <= 2*10⁹) that represent 
a node (i, j). You can assume that this is a valid node in the binary tree described above.

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing two numbers l and r separated by a single space, where l is how often you have to go left and r is how often you have to go right when traversing the tree from the root to the node given in the input. Print an empty line after every scenario.

Sample Input

3
42 1
3 4
17 73

Sample Output

Scenario #1:
41 0

Scenario #2:
2 1

Scenario #3:
4 6

 1 /*
 2 題意：根（a,b），則左孩子為（a+b,b），右孩子為（a,a+b）
 3     給定（m,n），初試根為（1,1）,從（1,1）到（m,n）需要往左子樹走幾次，
 4     往右子樹走幾次。
 5 思路：逆向思維，從（m,n）到（1,1）。
 6     給定（m,n），求其父親，如果m>n，則他父親是（m-n,n），
 7     否則（m,n-m）。
 8 程式碼一：  ----TLE
 9 #include <iostream>
10 
11 using namespace std;
12 
13 int main()
14 {
15     int T, a, b, lcnt, rcnt;
16     cin >> T;
17     for(int i = 1; i <= T; ++i)
18     {
19         cin >> a >> b;
20         lcnt = rcnt = 0;
21         while(a != 1 || b != 1)
22         {
23             if(a >= b)
24             {
25                 ++lcnt;
26                 a -= b;
27             }
28             else
29             {
30                 ++rcnt;
31                 b -= a;
32             }
33         }
34         cout << "Scenario #" << i << ':' << endl;
35         cout << lcnt << ' ' << rcnt << endl <<endl;
36     }
37     return 0;
38 }
39 
40 程式碼二：
41 用除法代替減法，得到的商即為往左走的次數，最後的m=m%n。
42 n>m時情況類推。
43 需要特別注意的是：
44     如果m>n,m%n == 0 怎麼辦？因為根（1,1）不可能有0存在，所以特殊處理一下：
45 次數：m/n-1；m=1
46 */
47 #include <iostream>
48 
49 using namespace std;
50 
51 int main()
52 {
53     int T, a, b, lcnt, rcnt;
54     cin >> T;
55     for(int i = 1; i <= T; ++i)
56     {
57         cin >> a >> b;
58         lcnt = rcnt = 0;
59         while(a != 1 || b != 1)
60         {
61             if(a >= b)
62             {
63                 if(a % b)
64                 {
65                     lcnt += a/b;
66                     a %= b;
67                 }
68                 else
69                 {
70                     lcnt += a/b-1;
71                     a = 1;
72                 }
73             }
74             else
75             {
76                 if(b % a)
77                 {
78                     rcnt += b/a;
79                     b %= a;
80                 }
81                 else
82                 {
83                     rcnt += b/a-1;
84                     b = 1;
85                 }
86             }
87         }
88         cout << "Scenario #" << i << ':' << endl;
89         cout << lcnt << ' ' << rcnt << endl <<endl;
90     }
91     return 0;
92 }