#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pli = pair<ll, int>;
const int INF = 0x3f3f3f3f, N = 2e5 + 5;
const ll LINF = 1e18 + 5;
const ll up = 1e6;
ll n, k, a[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  if (k >= up) return cout << -1, 0;
  ll sum = (k + 1) * k / 2;
  if (sum > n) return cout << -1, 0;
  ll g = 1, d = n - sum;
  for (ll i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      if (i <= n / sum && i > g) {
        g = i;
        d = n - g * sum;
      }
      if (n / i <= n / sum && n / i > g) {
        g = n / i;
        d = n - g * sum;
      }
    }
  }
  if (n - n * sum == 0) {
    g = n;
    d = 0;
  }
  for (int i = 1; i <= k; i++) a[i] = g * i;
  a[k] += d;
  for (int i = 1; i <= k; i++) cout << a[i] << " \n"[i == n];
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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;
using ll = long long;
using pii = pair<int, int>;
using pli = pair<ll, int>;
const int INF = 0x3f3f3f3f, N = 2e5 + 5;
const ll LINF = 1e18 + 5;
const ll up = 1e6;
ll n, k, a[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  if (k >= up) return cout << -1, 0;
  ll sum = (k + 1) * k / 2;
  if (sum > n) return cout << -1, 0;
  ll g = 1, d = n - sum;
  for (ll i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      if (i <= n / sum && i > g) {
        g = i;
        d = n - g * sum;
      }
      if (n / i <= n / sum && n / i > g) {
        g = n / i;
        d = n - g * sum;
      }
    }
  }
  if (n - n * sum == 0) {
    g = n;
    d = 0;
  }
  for (int i = 1; i <= k; i++) a[i] = g * i;
  a[k] += d;
  for (int i = 1; i <= k; i++) cout << a[i] << " \n"[i == n];
  return 0;
}

```