Codeforces Round #287 (Div. 2) C. Guess Your Way Out! 思路

C. Guess Your Way Out!

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height h. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.

Let's index all the leaf nodes from the left to the right from 1 to 2^h. The exit is located at some node n where 1 ≤ n ≤ 2^h, the player doesn't know where the exit is so he has to guess his way out!

Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules:

• Character 'L' means "go to the left child of the current node";
• Character 'R' means "go to the right child of the current node";
• If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node;
• If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command;
• If he reached a leaf node that is not the exit, he returns to the parent of the current node;
• If he reaches an exit, the game is finished.

Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit?

Input

Input consists of two integers h, n (1 ≤ h ≤ 50, 1 ≤ n ≤ 2^h).

Output

Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.

Examples

Input

1 2

Output

2

Input

2 3

Output

5

Input

3 6

Output

10

Input

10 1024

Output

2046

Note

A perfect binary tree of height h is a binary tree consisting of h + 1 levels. Level 0 consists of a single node called root, level h consists of 2^h nodes called leaves. Each node that is not a leaf has exactly two children, left and right one.

Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit.

题意：给你一个满二叉树高度为h+1。每次往左，在往右，往左，求到叶子的从左往右数第n个的叶子，需要做过几个点；

思路：仔细观察你会发现，如果这个点的遍历值为偶数，必然往右走，同理，奇数，往左走；

　　　显然你只要将n点在当前区间左右，如果下面的点在左子树就加1，否则再加上子树的大小

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=2e5+10,M=4e6+10,inf=1e9+10,mod=1e9+7;
const ll INF=1e18+10;
int main()
{
    ll h,n;
    scanf("%lld%lld",&h,&n);
    ll st=1;
    ll l=1;
    ll r=(1ll<<h);
    ll ans=0;
    while(1)
    {
        if(n==l&&n==r)break;
        ll mid=(l+r)>>1;
        if(n<=mid)
        {
            if(st&1)
                st+=1;
            else
                st+=(1ll<<(h+1)-1);
            r=mid;
        }
        else
        {
            if(st&1)
                st+=(1ll<<(h+1)-1);
            else
                st+=1;
            l=mid+1;
        }
        h--;
    }
    printf("%lld
",st-1);
    return 0;
}