Codeforces 862C. Mahmoud and Ehab and the xor

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.

容易想到a xor a = 0等特殊性质，关键是怎么凑出n个数

以下是tutorial中内容：
  取pw = 2^17 

  首先排除特殊情况，n=1,n=2，和唯一为NO的特例n=2,x=0

  然后首先计算出1-n-3的异或和=y，然后如果y=x,则再添加pw,pw*2,(pw xor (pw*2)),否则添加pw,0,pw xor x xor y

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

#define SIZE 100005

LL pw,n,x;
int main(){
  // freopen("test.in","r",stdin);
  cin >> n >> x;
  pw = pow(2,17);
  if (n == 1){
    cout << "YES" << endl << x; return 0;
  }
  if (n == 2){
    if (x == 0){
      cout << "NO";
    }
    else {
      cout << "YES" << endl;
      cout << "0 " << x;
    }
    return 0;
  }
  cout << "YES" << endl;

  LL y = 0;
  for (int i=1;i<=n-3;i++){
    y = y ^ i;
    cout << i << " ";
  }
  if (y == x){
    cout << pw << " " << pw*2 << " " << (pw^(pw*2));
  }
  else {
    cout << pw << " 0 " << (pw^x^y);
  }

  return 0;
}