模拟——Educational Codeforces Round 11——B

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：有2n排座位，n排在左边，n排在右边。有m个人（编号为1-m）依次上车，优先坐左边靠窗的座位，再是右边靠窗的座位，然后是左边不靠窗，最后是右边不靠窗。要输出下车时人的循序编号。
下车时左边第一排不靠窗先下，然后是靠窗，再是右边第一排不靠窗下，然后靠窗，每一排依次下车直到所有人下车。
题解:模拟即可。

/*B*/
#include<cstdio>
#include<cstring>
using namespace std;

int main()
{
    int n,m;
    int seat[5][110];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(seat,0,sizeof(seat));
        seat[1][1]=1;seat[4][1]=2;
        for(int i=2;i<=n;i++)
        {
            seat[1][i]=seat[1][i-1]+2;
            seat[4][i]=seat[4][i-1]+2;
        }
        seat[2][1]=seat[4][n]+1;seat[3][1]=seat[2][1]+1;
        for(int i=2;i<=n;i++)
        {
            seat[2][i]=seat[2][i-1]+2;
            seat[3][i]=seat[3][i-1]+2;
        }
        for(int i=0;i<=4;i++)
            for(int j=0;j<=n;j++)
                if(seat[i][j]>m)
                    seat[i][j]=0;
        int s=0;
        for(int i=1;i<=n;i++)
        {
            if(seat[2][i])
            {
                printf("%d ",seat[2][i]);
                s++;
            }
            if(s==m)    break;
            if(seat[1][i])
            {
                printf("%d ",seat[1][i]);
                s++;
            }
            if(s==m)    break;
            if(seat[3][i])
            {
                printf("%d ",seat[3][i]);
                s++;
            }
            if(s==m)    break;
            if(seat[4][i])
            {
                printf("%d ",seat[4][i]);
                s++;
            }
            if(s==m)    break;
        }
        printf("
");
    }
    return 0;
}