Codeforces 862C

C. Mahmoud and Ehab and the xor

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

5 5

output

YES1 2 4 5 7

input

3 6

output

YES1 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 .

For the second sample .

题意理解：

将x分割为n个不同的整数

n == 1的时候直接输出 x

n==2 的时候我们可以输出 只要x ！= 0 我们可以输出 0 x

else {

随机输出n-3个数

最后三个数 a b 非常大的两个不相等的数 和一个(z ^ a^ b^x^(a[1]---a[n-3]))///这样确保异或结果的正确性。

这n个数的异或结果就是x

因为y^y^x=x;

}

#include<stdio.h>#include<algorithm>using namespace std;int a[1000005];int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        if(n==1)        {            printf("YES\n%d\n",m);        }        else if(n==2)        {            if(m==0)            {                printf("NO\n");            }            else            {                printf("YES\n0 %d\n",m);            }        }        else        {            int x=1<<18,y=1<<19,z=0;            printf("YES\n");            for(int i=1; i<=n-3; i++)            {                printf("%d ",i);                z=z^i;            }            printf("%d %d %d\n",x,y,(x^y^z^m));        }    }}