Educational Codeforces Round 11（B）模拟

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：看图就可以了。。不解释了。。。。

题解：我是找了下规律，按照x=（2*n+1）,y=1,x++,y++填充数组，输出是只要是<=m的都输出，就是答案啦

#include<cstdio>  #include<cstring>  #include<cstdlib>  #include<cmath>  #include<iostream>  #include<algorithm>  #include<vector>  #include<map>  #include<set>  #include<queue>  #include<string>  #include<bitset>  #include<utility>  #include<functional>  #include<iomanip>  #include<sstream>  #include<ctime>  using namespace std;#define N int(1e5)  #define inf int(0x3f3f3f3f)  #define mod int(1e9+7)  typedef long long LL;  int ans[N];int main(){#ifdef CDZSC  freopen("i.txt", "r", stdin);//freopen("o.txt","w",stdout);  int _time_jc = clock();#endif  int n, m;while (~scanf("%d%d", &n, &m)){int fst = 2 * n + 1;int sec = 1;memset(ans, 0, sizeof(ans));for (int i = 1; i <= n * 4; i+=2){ans[i] = fst++;ans[i + 1] = sec++;}for (int i = 1; i <= 4 * n; i++){if (ans[i] <= m){printf("%d ", ans[i]);}}puts("");}return 0;}