codeforces 755D. PolandBall and Polygon

D. PolandBall and Polygon

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

PolandBall has such a convex polygon with n veritces that no three of its diagonals intersect at the same point. PolandBall decided to improve it and draw some red segments.

He chose a number k such that gcd(n, k) = 1. Vertices of the polygon are numbered from 1 to n in a clockwise way. PolandBall repeats the following process n times, starting from the vertex 1:

Assume you've ended last operation in vertex x (consider x = 1 if it is the first operation). Draw a new segment from vertex x to k-th next vertex in clockwise direction. This is a vertexx + k or x + k - n depending on which of these is a valid index of polygon's vertex.

Your task is to calculate number of polygon's sections after each drawing. A section is a clear area inside the polygon bounded with drawn diagonals or the polygon's sides.

Input

There are only two numbers in the input: n and k (5 ≤ n ≤ 106, 2 ≤ k ≤ n - 2, gcd(n, k) = 1).

Output

You should print n values separated by spaces. The i-th value should represent number of polygon's sections after drawing first i lines.

Examples

input

5 2

output

2 3 5 8 11 

input

10 3

output

2 3 4 6 9 12 16 21 26 31 

Note

The greatest common divisor (gcd) of two integers a and b is the largest positive integer that divides both a and b without a remainder.

For the first sample testcase, you should output "2 3 5 8 11". Pictures below correspond to situations after drawing lines.

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题目大意：一个n边形，从1号点开始，每次走到(x+k-1)%n+1位置，问每次留下来的路径把这个多边形划分成了几个部分。2 ≤ k ≤ n - 2, gcd(n, k) = 1，可以发现一定每个点一定经过一次，插入边的时候对应的点对应了一个区间，两条直线不相交(贡献答案）当且仅当他们的区间无交集，因为一个点只对应一个区间(只考虑出去的那个)，用树状数组维护一下即可。

 有个细节：如果2k>n，那么把k转化一下为n-k就可以了。(以为这个细节很显然，然后就没去hack,结果本来房间里10个对的然后就只剩3个没fst，MDZZ)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstdlib>
#include<cmath>
#include<cstring>
using namespace std;
#define maxn 1000010
#define llg long long 
#define yyj(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
llg n,m,cs[maxn],k,x,y,ans,c[maxn];

llg lowbit(llg x) {return x&-x;}
void add(llg w,llg v)
{
    while (w<=n)
    {
        c[w]+=v; w+=lowbit(w);
    }
}
llg sum(llg w)
{
    llg ans=0;
    while (w>0)
        {
            ans+=c[w]; w-=lowbit(w);
        }
    return ans;
}

llg work(llg x,llg y)
{
    
    llg l=y,r=y+n-k-k;
    if (r>n)
    {
        r-=n;
        return sum(n)-sum(l-1)+sum(r);
    }
    else return sum(r)-sum(l-1);
}

int main()
{
    yyj("D");
    cin>>n>>k;
    if (k>n/2) k=n-k;
    x=1;
    ans=1;// add(1,1);
    for (llg i=1;i<=n;i++)
    {
        y=x+k;
        if (y>n) y-=n;
        ans+=i-work(x,y);
        add(x,1);
        x=y;
        printf("%lld ",ans);
    } 
    return 0;
}