time limit per test 2 seconds

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

YES
1 2 4 5 7

題意：構造一個含有n個元素的序列，使得每個值不同，並且所有的數的異或和等於x。

思路：直接從1開始構造。然後對最後一個和第一個進行特殊考慮。

#include "bits/stdc++.h"
using namespace std;
const int maxn = 1e6 + 10;
int ans[maxn];
int main(int argc, char const *argv[])
{
    int N, K;
    scanf("%d%d", &N, &K);
    int res = 0;
    if (N == 1) return printf("YES\n%d\n", K), 0;
    if (N == 2 && K == 0) return printf("NO\n"), 0;
    for (int i = 1; i < N; i++) {
        ans[i] = i; res ^= i;
    }
    ans[N] = res^K;
    if (ans[N] < N) {
        if (ans[N] == 1) ans[N - 1] |= (1<<19);
        else ans[1] |= (1<<19);
        ans[N] |= (1<<19);
    }
    printf("YES\n%d", ans[1]);
    for (int i = 2; i <= N; i++) {
        printf(" %d", ans[i]);
    }
    putchar(‘\n‘);
    return 0;
}

codeforces 862C. Mahmoud and Ehab and the xor