Codeforces D. Polandball and Polygon

D. Polandball and PolygonTime limit per test4 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

Polandball have such a convex polygon with n veritces that no three of it diagonals intersect at the same POI NT. 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 is numbered From 1 to  N  in a clockwise the. Polandball repeats the following process  N  times, starting from the vertex& nbsp 1:

Assume you ' ve ended last operation in vertex x (consider x = 1 If it's the first operation). Draw a new segment from vertex x to k-th next vertex in clockwise direction. This was a vertexx + k or x + k - n depending on which of t Hese 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 is 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.< /c18>

Examplesinput

5 2

Output

Input

10 3

Output

Note

The greatest common divisor (GCD) of the integers a and b are the largest positive integer that D Ivides 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.

The main topic: an N-sided shape, starting from point 1th, each walk to the (x+k-1)%n+1 position, ask each left the path to divide the polygon into several parts. 2 ≤ k ≤ n -2, gcd( n, k) = 1, it can be found that each point must pass one time, the corresponding point when inserting the edge corresponds to an interval, the two lines do not intersect ( Contribute to the answer) when and only if their interval is not intersection, because a point only corresponds to one interval (only consider the one that goes out), with a tree-like array maintenance can be.

There is a detail: if 2k>n, then the K conversion to n-k on it. (thought this detail is obvious, then did not go to the hack, the result originally 10 in the room is right then only then 3 not fst,mdzz)

1#include <iostream>2#include <cstdio>3#include <algorithm>4#include <vector>5#include <cstdlib>6#include <cmath>7#include <cstring>8 using namespacestd;9 #defineMAXN 1000010Ten #defineLLG Long Long One #defineYyj (a) freopen (a ".", "R", stdin), Freopen (a ". Out", "w", stdout); A Llg N,M,CS[MAXN],K,X,Y,ANS,C[MAXN]; -  -Llg lowbit (Llg x) {returnx&-x;} the voidAdd (llg w,llg v) - { -      while(w<=N) -     { +C[w]+=v; w+=Lowbit (w); -     } + } A llg sum (llg w) at { -LLG ans=0; -      while(w>0) -         { -ANS+=C[W]; w-=Lowbit (w); -         } in     returnans; - } to  + llg work (llg x,llg y) - { the      *LLG l=y,r=y+n-k-K; $     if(r>N)Panax Notoginseng     { -r-=N; the         returnSUM (n)-sum (l1)+sum (r); +     } A     Else returnSUM (r)-sum (l1); the } +  - intMain () $ { $Yyj"D"); -Cin>>n>>K; -     if(k>n/2) k=n-K; thex=1; -ans=1;//Add (+);Wuyi      for(Llg i=1; i<=n;i++) the     { -y=x+K; Wu         if(y>n) y-=N; -ans+=i-Work (x, y); AboutAdd (x,1); $x=y; -printf"%lld", ans); -     }  -     return 0; A}

Codeforces D. Polandball and Polygon