D. Ehab and the Expected XOR Problem

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本文介绍了一个算法问题，目标是构造一个数列，其中任意非空子段的位运算异或结果既不等于0也不等于给定值x。文章提供了解决方案的思路，即通过维护前缀异或和的特性来构建数列，并给出了一段C++代码实现。

output

standard output

Given two integers nn and xx, construct an array that satisfies the following conditions:

for any element aiai in the array, 1≤ai<2n1≤ai<2n;
there is no non-empty subsegment with bitwise XOR equal to 00 or xx,
its length ll should be maximized.

A sequence bb is a subsegment of a sequence aa if bb can be obtained from aa by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

Input

The only line contains two integers nn and xx (1≤n≤181≤n≤18, 1≤x<2181≤x<218).

Output

The first line should contain the length of the array ll.

If ll is positive, the second line should contain ll space-separated integers a1a1, a2a2, ……, alal (1≤ai<2n1≤ai<2n) — the elements of the array aa.

If there are multiple solutions, print any of them.

Examples

input

Copy

3 5

output

Copy

3
6 1 3

input

Copy

2 4

output

Copy

3
1 3 1

input

Copy

1 1

output

Copy

Note

In the first example, the bitwise XOR of the subsegments are {6,7,4,1,2,3}{6,7,4,1,2,3}.

题意：还是比较简单的，就是要你构造一组数列，所有子段和不为0或x，取值在2的次方内。

思路：利用好前缀异或和的性质即可。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<vector>
#include<algorithm>
#define LL long long

using namespace std;
const int maxn=3e5+100;
int vis[maxn];

int main()
{
    int n,x;
    cin>>n>>x;
    vis[0]=1;
    vector<int>st;
    st.push_back(0);
    for(int i=1;i<(1<<n);i++)
    {
        if(vis[i^x])
        {
            continue;
        }
        vis[i]=1;
        st.push_back(i);
    }
    cout<<st.size()-1<<endl;
    for(int i=1;i<st.size();i++)
    {
        cout<<(st[i]^st[i-1])<<" ";
    }
    cout<<endl;
    return 0;
}