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 <bits/stdc++.h>
using namespace std;
int main()
{
    int N, W;
    cin >> N >> W;
    const int H = W / 2;
    vector<int> imos(N + 1, 0);
    imos[1] = H;
    for (int j = H + 1; j < N + W; j++) {
        imos[max(1, j - W + 1)]++;
        if (j + 1 <= N) { imos[j + 1]--; }
    }
    for (int j = H + 1; j < W; j++) {
        for (int k = 1; k * j + 2 * j - W + 1 <= N; k++) {
            imos[k * j + 2 * j - W + 1]++;
            if (k * j + j + 1 <= N) { imos[k * j + j + 1]--; }
        }
    }
    for (int i = 1; i <= N - W + 1; i++) { imos[i] += imos[i - 1], cout << (i == 1 ? "" : " ") << imos[i]; }
    cout << endl;
    return 0;
}

