Mahmoud and Ehab and the xor CodeForces - 862C （异或性质） 4000

Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won’t show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.

Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn’t like big numbers, so any number in the set shouldn’t be greater than 106.

Input

The only line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the set and the desired bitwise-xor, respectively.

Output

If there is no such set, print “NO” (without quotes).

Otherwise, on the first line print “YES” (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.

Example

Input

5 5

Output

YES

1 2 4 5 7

Input

3 6

Output

YES

1 2 5

Note

You can read more about the bitwise-xor operation here: https://en.wikipedia.org/wiki/Bitwise_operation#XOR

For the first sample .

1^2^4^5^7=5

For the second sample .

1^2^5=6

大致题意：告诉你n和x，问你能不能找出n个不同的数使得他们的异或值为x。

思路：由异或的性质可知，当两个数一样时，它们的异或值为0，当0异或上一个数x时的值为x。所以当n为2，x为0时，我们无法找到不同的两个数使得它们的异或值为0，这是唯一的一种输出NO的情况，当n为1时，直接输出x，当n大于等于三时，我们假设ans为前1到n-3的所有数和x的异或值，a为2^17,b为2^18，那么，最后的三个数即为ans^a，a^b，b。

代码如下

#include<cstring>
#include<cstdio>
#include<iostream>
#include <algorithm>
using namespace std;
const int num=131072;
int main()
{

int n,x;
cin>>n>>x;
if(n==2)
{
if(x==0)
cout<<"NO";
else
cout<<"YES\n",cout<<num<<' '<<(num^x);
}
else if(n==1)
cout<<"YES\n",cout<<x;
else
{
cout<<"YES\n";
for(int i=1;i<=n-3;i++)
cout<<i<<' ',x^=i;
cout<<(x^num)<<' '<<2*num<<' '<<(2*num^(num));
}
return 0;
}