Codeforces 743B-Chloe and the sequence

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.

题意：

1初始化

1 2 1置入 2

1 2 1 3 1  2 1置入 3

1 2 1 3 1  2 1 4 1 2 1 3 1  2 1置入 4

1 2 1 3 1  2 1 4 1 2 1 3 1  2 1 5 1 2 1 3 1  2 1 4 1 2 1 3 1  2 1置入 5

1 2 1 3 1  2 1 4 1 2 1 3 1  2 1 5 1 2 1 3 1  2 1 4 1 2 1 3 1  2 1 6 1 2 1 3 1  2 1 4 1 2 1 3 1  2 1 5 1 2 1 3 1  2 1 4 1 2 1 3 1  2 1置入 6

给定一个整数 n ，操作 n-1 次后，问第 k 位是多少。操作如上。

解题思路：递归求解

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <stack>#include <queue>#include <vector>#include <map>using namespace std;#define ll long longll mypow(ll x,ll y){    ll sum=1;    while(y)    {        if(y&1) sum*=x;        y/=2;        x*=x;    }    return sum;}int main(){    ll n,k;    while(~scanf("%lld %lld",&n,&k))    {        ll i,x;        for(i=n;;i--)        {            x=mypow(2,i-1);            if(k>x) k=x-(k-x);            else if(k==x) break;            if(k==1||k==2) break;        }        if(k==1||k==2) printf("%lld\n",k);        else printf("%lld\n",i);    }    return 0;}