#include <bits/stdc++.h>
using namespace std;
const long long MX = 202020;
const long long INF = 9e18;
long long n, k, ans;
long long fun(long long x) { return k * (k + 1) / 2 <= n / x; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  if (k > 200000) {
    cout << -1;
    return 0;
  }
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      if (fun(i)) ans = max(ans, i);
      if (i != n / i && fun(n / i)) ans = max(ans, n / i);
    }
  }
  if (ans == 0) {
    cout << -1;
    return 0;
  }
  n /= ans;
  for (int i = 1; i <= k - 1; i++) {
    cout << i * ans << " ";
    n -= i;
  }
  cout << n * ans;
}

### Prompt

Develop a solution in cpp to the problem described below:

You are given positive integer number n. You should create such strictly increasing sequence of k positive numbers a1, a2, ..., ak, that their sum is equal to n and greatest common divisor is maximal.

Greatest common divisor of sequence is maximum of such numbers that every element of sequence is divisible by them.

If there is no possible sequence then output -1.

Input

The first line consists of two numbers n and k (1 ≤ n, k ≤ 1010).

Output

If the answer exists then output k numbers — resulting sequence. Otherwise output -1. If there are multiple answers, print any of them.

Examples

Input

6 3


Output

1 2 3


Input

8 2


Output

2 6


Input

5 3


Output

-1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MX = 202020;
const long long INF = 9e18;
long long n, k, ans;
long long fun(long long x) { return k * (k + 1) / 2 <= n / x; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  if (k > 200000) {
    cout << -1;
    return 0;
  }
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      if (fun(i)) ans = max(ans, i);
      if (i != n / i && fun(n / i)) ans = max(ans, n / i);
    }
  }
  if (ans == 0) {
    cout << -1;
    return 0;
  }
  n /= ans;
  for (int i = 1; i <= k - 1; i++) {
    cout << i * ans << " ";
    n -= i;
  }
  cout << n * ans;
}

```