Solve in Python:
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 vertex x + 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 ≤ 10^6, 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. [Image] [Image] [Image] [Image] [Image] [Image]
def __starting_point():
    n,k = list(map(int, input().strip().split()))

    if k > n//2:
        k = n - k

    intersection = n * [0]

    count = 1

    done = False
    i=0
    result = []
    for i in range(1,n+1):
        nn = (i*k) // n
        j = (i*k)%n
        if j < k:
            count += (2*nn )
        else:
            count += (2*nn +1)
        result.append(count)

    result[-1] -= 1
    print(" ".join([str(r) for r in result]))

__starting_point()