B. After Training

B. After Training

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 hasn balls andm baskets. The baskets are positioned in a row from left to right and they are numbered with numbers from1 tom, correspondingly. The balls are numbered with numbers from1 ton.

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

2132

Input

3 1

Output

111
//超时的代码：O(n*n*logn)#include <iostream>#include <cstdio>#include <vector>#include <map>#include <set>#include <string>#include <cstring>#include <cstdlib>#include <cmath>#include <utility>#include <algorithm>using namespace std;const int maxn = 1e5+10;int n, m;struct bas{    int num;    int balln;};bool cmp(const bas & a, const bas & b){    double x = (m+1)/2.0-a.num;    double y = (m+1)/2.0-b.num;    if (x < 0)    {        x = -x;    }    if (y < 0)    {        y = -y;    }    if (a.balln != b.balln)    {        return a.balln < b.balln;    }    else if (x != y)    {        return x < y;    }    else    {        return a.num < b.num;    }}bas b[maxn];int main(){    while (scanf("%d%d", &n, &m) != EOF)    {        for (int i=1; i<=m; i++)        {            b[i].num = i;            b[i].balln = 0;        }        while (n--)        {            sort(b+1, b+m+1, cmp);            b[1].balln++;            cout << b[1].num << endl;        }    }    return 0;}

//AC代码：O(n*logn)#include <iostream>#include <cstdio>#include <vector>#include <map>#include <set>#include <string>#include <cstring>#include <cstdlib>#include <cmath>#include <utility>#include <algorithm>using namespace std;const int maxn = 1e5+10;int n, m;struct bas{    int num;    int balln;};struct rule{    bool operator()(const bas & a, const bas & b)    {        double x = (m+1)/2.0-a.num;        double y = (m+1)/2.0-b.num;        if (x < 0)        {            x = -x;        }        if (y < 0)        {            y = -y;        }        if (a.balln != b.balln)        {            return a.balln < b.balln;        }        else if (x != y)        {            return x < y;        }        else        {            return a.num < b.num;        }    }};//bas b[maxn];set<bas, rule> b;set<bas, rule> :: iterator it;int main(){    bas ba;    scanf("%d%d", &n, &m);    for (int i=1; i<=m; i++)    {        ba.num = i;        ba.balln = 0;        b.insert (ba);    }    while (n--)    {        ba = *b.begin();        ba.balln++;        cout << ba.num << endl;        b.erase(b.begin());        b.insert (ba);    }    return 0;}

终于又有一次这样的开心O(∩_∩)O~~，，，独自解决问题的成就感，哦呵呵~