B. Seating On Bus

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

这个题目挺简单，我是分情况讨论的。

#include <iostream>#include <cstdio>using namespace std;int main(){    int n,m,i;    cin>>n>>m;    if(m>2*n){        for(i=1;i<=2*n;){            if(m>1){                printf("%d %d ",i+2*n,i);                m--;                i++;            }            if(m<=2*n)break;            if(m>1){                printf("%d %d ",i+2*n,i);                m--;                i++;            }            if(m<=2*n)break;        }        for(;i<=2*n;i++)            printf("%d ",i);        printf("\n");    }    else{        for(i=1;i<m;i++)            printf("%d ",i);        printf("%d\n",i);    }    return 0;}

看了下排行的大神的代码，真的是感受到了思维的差距，明明很简单就可以解决，我缺写了那么多，加油啊！

# include <cstdio>int main(){int n, m;scanf("%d%d", &n, &m);for (int i = 1; i <= n * 2; ++i){if (i + n * 2 <= m)printf("%d ", i + n * 2);if (i <= m)printf("%d ", i);}}