Codeforces Round #384 (Div. 2) B. Chloe and the sequence

传送门：http://codeforces.com/contest/743/problem/B

B. Chloe and the sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.

Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (n - 1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1, 2, 1] after the first step, the sequence [1, 2, 1, 3, 1, 2, 1] after the second step.

The task is to find the value of the element with index k (the elements are numbered from 1) in the obtained sequence, i. e. after (n - 1)steps.

Please help Chloe to solve the problem!

Input

The only line contains two integers n and k (1 ≤ n ≤ 50, 1 ≤ k ≤ 2n - 1).

Output

Print single integer — the integer at the k-th position in the obtained sequence.

Examples

input

3 2

output

2

input

4 8

output

4

Note

In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.

In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.

这就是一道递归题：

下面是错误代码，在处理数据时发生了溢出，在晚上提交的时候，一次就过了，但是在第二天就被hack了

#include <iostream>
#include <cmath>
using namespace std;

long long  solve(long long n,long long k){
    if(k>1<<n-1)  //在移位的时候发生了溢出，1默认是int型的，应该强制转换一下的
       k=k%(1<<n-1);  

    if(k%(1<<(n-1))==0)
    return  n;
    solve(n-1,k);
}
  int main(){
    long long n,k;
    cin>>n>>k;
    cout<<solve(n,k)<<endl;
}

下面的是正确的代码：

#include <iostream>
#include <cmath>
using namespace std;

const long long S=1;
long long  solve(long long n,long long k){
    if(k>S<<n-1)
       k=k%(S<<n-1);

    if(k%(S<<(n-1))==0)
    return  n;
    solve(n-1,k);
}


int main(){
    long long n,k;
    cin>>n>>k;
    cout<<solve(n,k)<<endl;
}