B. Seating On Bus

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

我的：这道题刚开始的时候感觉就是一个水题，但是第一次提交之后就WA了，之后感觉补了好多次，从比赛开始三十分钟到距离结束前几分钟都在不漏洞，最后终于AC了，汗呀！

接下来是我写的代码：

#include<iostream>
using namespace std;
const int maxn=100+5;
int main()
{
int a[maxn][2],b[maxn][2];
int n,m;
while(cin>>n>>m)
{
int x=0,y=0;
int t=1;
int i;
for(i=0;i<n;i++)
{
a[i][0]=0;a[i][1]=0;
b[i][0]=0;b[i][1]=0;
}
if(m>=2*n)
{
while(t<=2*n)
{
a[x++][0]=t++;
if(t>2*n)break;
b[y++][1]=t++;
}
//cout<<b[y-1][1]<<endl;
x=0;y=0;
while(t<=m)
{
a[x++][1]=t++;
if(t>m)break;
b[y++][0]=t++;
}
}
else
{
t=1;
x=0;y=0;
while(t<=m)
{
a[x++][0]=t++;
if(t>m)break;
b[y++][1]=t++;
}
}

for(i=0;i<n;i++)
if(i!=n-1)
{
if(a[i][1])cout<<a[i][1]<<" ";
if(a[i][0])cout<<a[i][0]<<" ";
if(b[i][0])cout<<b[i][0]<<" ";
if(b[i][1])cout<<b[i][1]<<" ";
}
else
{
if(a[i][1])cout<<a[i][1]<<" ";
if(a[i][0])cout<<a[i][0]<<" ";
if(b[i][0])cout<<b[i][0]<<" ";
if(b[i][1])cout<<b[i][1]<<endl;
}
}
return 0;
}