Codeforces Gym100952 B. New Job (2015 HIAST Collegiate Programming Contest)

 

B. New Job

 

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

This is the first day for you at your new job and your boss asks you to copy some files from one computer to other computers in an informatics laboratory. He wants you to finish this task as fast as possible. You can copy the files from one computer to another using only one Ethernet cable. Bear in mind that any File-copying process takes one hour, and you can do more than one copying process at a time as long as you have enough cables. But you can connect any computer to one computer only at the same time. At the beginning, the files are on one computer (other than the computers you want to copy them to) and you want to copy files to all computers using a limited number of cables.

Input

First line of the input file contains an integer T (1  ≤  T  ≤  100) which denotes number of test cases. Each line in the next T lines represents one test case and contains two integers N, M.

N is the number of computers you want to copy files to them (1  ≤  N  ≤  1,000,000,000). While M is the number of cables you can use in the copying process (1  ≤  M  ≤  1,000,000,000).

Output

For each test case, print one line contains one integer referring to the minimum hours you need to finish copying process to all computers.

Examples

Input

3
10 10
7 2
5 3

Output

4
4
3

Note

In the first test case there are 10 computer and 10 cables. The answer is 4 because in the first hour you can copy files only to 1 computer, while in the second hour you can copy files to 2 computers. In the third hour you can copy files to 4 computers and you need the fourth hour to copy files to the remaining 3 computers.

 

题意就是复印文件，电缆有限，然后就是在电缆数允许的情况下1台电脑复印之后就有2台电脑可以继续进行复印，然后就是4台。。。达到电缆数所允许的最大值的时候剩下的就是按电缆数进行复印。

ヾ(｡｀Д´｡)，写这个题的时候智商不在家，wa了好几次。

 

代码:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     int t,n,m,i,cnt,ans;
 5     scanf("%d",&t);
 6     while(t--){
 7         scanf("%d%d",&n,&m);
 8         cnt=1;
 9         for(i=0;n>0;i++){
10             if(cnt>m)n-=m;
11             else{
12                n-=cnt;
13                cnt*=2;
14             }
15         }
16         printf("%d\n",i);
17     }
18     return 0;
19 }