CodeForces - 195B (模拟)

After a team finished their training session on Euro football championship, Valeric was commissioned to gather the balls and sort them into baskets. Overall the stadium has n balls and m baskets. The baskets are positioned in a row from left to right and they are numbered with numbers from 1 to m, correspondingly. The balls are numbered with numbers from 1 to n.

Valeric decided to sort the balls in the order of increasing of their numbers by the following scheme. He will put each new ball in the basket with the least number of balls. And if he's got several variants, he chooses the basket which stands closer to the middle. That means that he chooses the basket for which  is minimum, where i is the number of the basket. If in this case Valeric still has multiple variants, he chooses the basket with the minimum number.

For every ball print the number of the basket where it will go according to Valeric's scheme.

Note that the balls are sorted into baskets in the order of increasing numbers, that is, the first ball goes first, then goes the second ball and so on.

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 10⁵) — the number of balls and baskets, correspondingly.

Output

Print n numbers, one per line. The i-th line must contain the number of the basket for the i-th ball.

Examples

Input

4 3

Output

2
1
3
2

Input

3 1

Output

1
1
1

题目给出 n 个球和 m 个筐子，装法是第一个球放中间，然后第二个球放前面，第三个球放后面，以此类推，偶数个筐子的话就是前两个球放在中间。
若数量相同再放距离中间篮子最近的，若距离相同放篮子编号小的。n可能是大于m的，所以我们先把前m个球排好，然后后面的重复前面的操作就行了，题目要求输出放入球的篮子的顺序

#include <cstdio>
#include <cstring>
using namespace std;
const int N = 1e5 + 5;
int a[N];

int main()
{
    int n, m;
    while(~scanf("%d%d", &n, &m)) {
        if(m & 1)  //判断是否为奇数，因为n为奇数时，对应的二进制数最低位一定为1，n&1的结果就是1
        {
            int mid = m / 2 + 1;  
            a[1] = mid;
            for(int i = 2, k = 1; i <= m; i++) {
                if(i & 1) {
                    a[i] = mid + k;
                    k++;
                }
                else    
					a[i] = mid - k;
            }
        }
        else {
            int mid = m / 2;
            a[1] = mid;  a[2] = mid + 1;
            for(int i = 3, k = 1; i <= m; i++) {
                if(i & 1)
                    a[i] = mid - k;
                else {
                    a[i] = mid + 1 + k;
                    k++;
                }
            }
        }
        for(int i = m + 1, k = 1; i <= n; i++) {
            a[i] = a[k];
            if(k == m)
                k = 1;
            else
                k++;
        }
        for(int i = 1; i <= n; i++)
            printf("%d\n", a[i]);
    }
    return 0;
}