Sliding GCD

Problem Statement

For the set S of natural numbers, let f (S) be the number of elements in the set \\ {GCD (T) | T ⊆ S, T is not empty \\}.

Here, GCD (T) is the greatest integer that divides all the numbers contained in T.
In particular, note that GCD (\\ {a \\}) = a when T consists of only one integer a.

Find f (\\ {i, i + 1, ..., i + W --1 \\}) for i = 1, 2, ..., N --W + 1.

Constraints

* 1 ≤ W ≤ N ≤ 10 ^ 5

Input

Input follows the following format. All given numbers are integers.


N W

Output

The value of f (\\ {i, i + 1, ..., i + W-1 \\}) when i = 1, 2, ..., N-W + 1 is separated by a space. Print on a line.

Examples

Input

10 2


Output

2 3 3 3 3 3 3 3 3


Input

30 7


Output

7 8 9 10 10 11 11 11 11 12 11 12 10 12 12 11 10 12 12 12 10 11 11 13
#include <iostream>
#include <algorithm>
using namespace std;
int N, W, s[200009];
int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);
	cin >> N >> W;
	int ret = 0;
	for (int i = W / 2 + 1; i < W; i++) {
		for (int j = 2 * i - W + 1, k = 1; j <= N - W + 1 || k == 1; j += i, k++) {
			int l = j, r = j + W - i;
			if (j == 2 * i - W + 1) l = 1;
			s[l]++, s[r]--;
		}
	}
	for (int i = 1; i <= N - W + 1; i++) {
		if (i >= 2) cout << ' ';
		s[i] += s[i - 1];
		cout << s[i] + W / 2 + min(i, W);
	}
	cout << "\n";
	return 0;
}
