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>
#include <vector>
using namespace std;

const int MAXN = 100005;
int N, W;
int num[MAXN];
int ans;
vector<int> d[MAXN];

void add(int v, int end) {
  for(int j = 0; j < d[v].size(); ++j) {
    int i = d[v][j];
    ++num[i];
    if(i < end && num[i] == 2) ++ans;
  }
}

void subtract(int v, int end) {
  for(int j = 0; j < d[v].size(); ++j) {
    int i = d[v][j];
    --num[i];
    if(i < end && num[i] == 1) --ans;
  }
}

int main() {
  for(int i = 1; i < MAXN; ++i) {
    for(int j = i; j < MAXN; j += i) {
      d[j].push_back(i);
    }
  }

  while(cin >> N >> W) {
    fill(num, num+MAXN, 0);
    ans = W;
    for(int i = 1; i < W; ++i) add(i, 1);
    for(int i = W; i <= N; ++i) {
      add(i, i-W+1);
      if(i != W) cout << " ";
      cout << ans;
      subtract(i-W+1, i-W+1);
      if(num[i-W+1] >= 2) {
        ++ans;
      }
    }
    cout << endl;
  }
  return 0;
}