codeforces 660B B. Seating On Bus(模拟)

题目链接：

B. Seating On Bus

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Consider 2n rows of the seats in a bus. n rows of the seats on the left and n rows of the seats on the right. Each row can be filled by two people. So the total capacity of the bus is 4n.

Consider that m (m ≤ 4n) people occupy the seats in the bus. The passengers entering the bus are numbered from 1 to m (in the order of their entering the bus). The pattern of the seat occupation is as below:

1-st row left window seat, 1-st row right window seat, 2-nd row left window seat, 2-nd row right window seat, ... , n-th row left window seat, n-th row right window seat.

After occupying all the window seats (for m > 2n) the non-window seats are occupied:

1-st row left non-window seat, 1-st row right non-window seat, ... , n-th row left non-window seat, n-th row right non-window seat.

All the passengers go to a single final destination. In the final destination, the passengers get off in the given order.

1-st row left non-window seat, 1-st row left window seat, 1-st row right non-window seat, 1-st row right window seat, ... , n-th row left non-window seat, n-th row left window seat, n-th row right non-window seat, n-th row right window seat.

The seating for n = 9 and m = 36.

You are given the values n and m. Output m numbers from 1 to m, the order in which the passengers will get off the bus.

 

Input

The only line contains two integers, n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 4n) — the number of pairs of rows and the number of passengers.

 

Output

Print m distinct integers from 1 to m — the order in which the passengers will get off the bus.

 

Examples

input

2 7

output

5 1 6 2 7 3 4

input

9 36

output

19 1 20 2 21 3 22 4 23 5 24 6 25 7 26 8 27 9 28 10 29 11 30 12 31 13 32 14 33 15 34 16 35 17 36 18

题意：

给一个车的状态,问乘客最后下车的次序;

思路：

用队列模拟一下啦;

AC代码：

/*
2014300227    660B - 6    GNU C++11    Accepted    31 ms    2176 KB
*/
#include <bits/stdc++.h>
using namespace std;
const int N=1e5+4;
typedef long long ll;
const double PI=acos(-1.0);
queue<int>qu1,qu2,qu3,qu4;
int n,m,num1,num2,num3,num4;
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m&&i<=2*n; )
    {
        if(i<=m&&num1<=n){
        qu1.push(i);
        num1++;
        i++;
        }
        if(i<=m&&num2<=n)
        {
            qu2.push(i);
            num2++;
            i++;
        }
    }
    for(int i=2*n+1;i<=m;)
    {
       if(i<=m){ qu3.push(i);
        i++;}

        if(i<=m){qu4.push(i);
        i++;}
    }
    while(m)
    {
        if(!qu3.empty())
        {
            printf("%d ",qu3.front());
            qu3.pop();
            m--;
        }
        if(!qu1.empty())
        {
            printf("%d ",qu1.front());
            qu1.pop();
            m--;
        }
        if(!qu4.empty())
        {
            printf("%d ",qu4.front());
            qu4.pop();
            m--;
        }
        if(!qu2.empty())
        {
            printf("%d ",qu2.front());
            qu2.pop();
            m--;
        }
    }


    return 0;
}