The Old Frog King lives on the root of an infinite tree. According to the law, each node should connect to exactly two nodes on the next level, forming a full binary tree. 

Since the king is professional in math, he sets a number to each node. Specifically, the root of the tree, where the King lives, is 11. Say froot=1froot=1. 

And for each node uu, labels as fufu, the left child is fu×2fu×2 and right child is fu×2+1fu×2+1. The king looks at his tree kingdom, and feels satisfied. 

Time flies, and the frog king gets sick. According to the old dark magic, there is a way for the king to live for another NN years, only if he could collect exactly NNsoul gems. 

Initially the king has zero soul gems, and he is now at the root. He will walk down, choosing left or right child to continue. Each time at node xx, the number at the node is fxfx (remember froot=1froot=1), he can choose to increase his number of soul gem by fxfx, or decrease it by fxfx. 

He will walk from the root, visit exactly KK nodes (including the root), and do the increasement or decreasement as told. If at last the number is NN, then he will succeed. 

Noting as the soul gem is some kind of magic, the number of soul gems the king has could be negative. 

Given NN, KK, help the King find a way to collect exactly NN soul gems by visiting exactly KK nodes.

Input

First line contains an integer TT, which indicates the number of test cases. 

Every test case contains two integers NN and KK, which indicates soul gems the frog king want to collect and number of nodes he can visit. 

⋅⋅ 1≤T≤1001≤T≤100. 

⋅⋅ 1≤N≤1091≤N≤109. 

⋅⋅ N≤2K≤260N≤2K≤260.

Output

For every test case, you should output " Case #x:

" first, where xx indicates the case number and counts from 11. 

Then KK lines follows, each line is formated as 'a b', where aa is node label of the node the frog visited, and bb is either '+' or '-' which means he increases / decreases his number by aa. 

It's guaranteed that there are at least one solution and if there are more than one solutions, you can output any of them. 
 

Sample Input

2
5 3
10 4

Sample Output

Case #1:
1 +
3 -
7 +
Case #2:
1 +
3 +
6 -
12 +

       這是一道思維題差不多的吧，一個向下延伸的二叉樹，問能否在b步的限制下獲得a的總價值~~；

       我們可以注意得到的是2^b>=a,也就是說b與二進位制下的a有關~~

       於是我們可以從常理上推得到，只要從1 2 4 8 16這樣的從最左邊走，一定能夠在n（2^n剛好>=a）的步數下到達。如果a是2的冪的話，最後一步只要走右邊那條路就好了。

        而如果多餘的步數我們可以吧最後一步的（+）改為減（可以得出一定是+），然後向這個最後一步的兩倍的下方延伸。

比如最後一步是+15，我們可以把多餘的步數（假如多3步）改為-15 -30 -60 +120.這樣的話，就能完美抵消多餘步數的影響了。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
#define ll long long int
ll a,b;
ll su[70];
bool check(){
    su[0]=1;
    for(int i=1;i<=62;i++){
        su[i]=su[i-1]*2;
    }
}
int main(){
    int te;
    cin>>te;
    check();
    for(int cas=1;cas<=te;cas++){
        printf("Case #%d:\n",cas);
        cin>>a>>b;
        if(a==1&&b==1){
            cout<<"1 +"<<'\n';
            continue;
        }
        int mi=lower_bound(su+0,su+62,a)-su;
      //  cout<<mi<<endl;
        if(su[mi]==a){
            queue<char>q;queue<ll>qu;
            ll now=1;
            for(int i=1;i<mi;i++){
                qu.push(now);
                q.push('+');
                now*=2;
            }
            for(int i=mi;i<b;i++){
                qu.push(now);
                q.push('-');
                now*=2;
            }
            qu.push(now+1);
            q.push('+');
            while(!q.empty()){
                cout<<qu.front()<<" ";
                cout<<q.front()<<"\n";
                qu.pop();
                q.pop();
            }
        }
        else{
            ll now=1;
            ll md=su[mi]-a;
      //      cout<<md<<endl;
            int spot=md%2;md/=2;
            queue<char>q;queue<ll>qu;
            for(int i=1;i<mi;i++){
                qu.push(now);
                if(md%2)q.push('-');
                else q.push('+');
                now*=2;
                md/=2;
            }
            if(spot==0)
                now+=1;
            for(int i=mi;i<b;i++){
                qu.push(now);
                q.push('-');
                now*=2;
            }
            qu.push(now);
            q.push('+');
            while(!q.empty()){
                cout<<qu.front()<<" ";
                cout<<q.front()<<"\n";
                qu.pop();
                q.pop();
            }
        }
    }
}