#include <bits/stdc++.h>
using namespace std;
int criba[100010 + 1];
void crearCriba() {
  int w[] = {4, 2, 4, 2, 4, 6, 2, 6};
  for (int p = 25; p <= 100010; p += 10) criba[p] = 5;
  for (int p = 9; p <= 100010; p += 6) criba[p] = 3;
  for (int p = 4; p <= 100010; p += 2) criba[p] = 2;
  for (int p = 7, cur = 0; p * p <= 100010; p += w[cur++ & 7])
    if (!criba[p])
      for (int j = p * p; j <= 100010; j += (p << 1))
        if (!criba[j]) criba[j] = p;
}
vector<int> primos;
void buscarPrimos() {
  crearCriba();
  for (int i = (2); i < (100010 + 1); i++)
    if (!criba[i]) primos.push_back(i);
}
void divisores(map<unsigned long long, unsigned long long> &f,
               vector<unsigned long long> &divs,
               map<unsigned long long, unsigned long long>::iterator it,
               unsigned long long n = 1) {
  if (it == f.begin()) divs.clear();
  if (it == f.end()) {
    divs.push_back(n);
    return;
  }
  unsigned long long p = it->first, k = it->second;
  ++it;
  for (int _ = (0); _ < (k + 1); _++) divisores(f, divs, it, n), n *= p;
}
void fact(unsigned long long n,
          map<unsigned long long, unsigned long long> &f) {
  for (auto p = primos.begin(); p != primos.end(); p++) {
    while (!(n % *p)) {
      f[*p]++;
      n /= *p;
    }
  }
  if (n > 1) f[n]++;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  buscarPrimos();
  unsigned long long n, k;
  cin >> n >> k;
  if (k > 1e6 || k * (k + 1) / 2 > n) {
    cout << "-1\n";
    return 0;
  }
  map<unsigned long long, unsigned long long> f;
  vector<unsigned long long> divs;
  fact(n, f);
  divisores(f, divs, f.begin());
  sort(divs.begin(), divs.end());
  int ans = 1;
  for (auto it = divs.begin(); it != divs.end(); it++) {
    if (*it * (k * (k + 1) / 2) <= n) {
      ans = *it;
    } else
      break;
  }
  unsigned long long acum = 0;
  for (int i = (1); i < (k); i++) {
    acum += ans * i;
    cout << ans * i << ' ';
  }
  cout << n - acum << '\n';
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int criba[100010 + 1];
void crearCriba() {
  int w[] = {4, 2, 4, 2, 4, 6, 2, 6};
  for (int p = 25; p <= 100010; p += 10) criba[p] = 5;
  for (int p = 9; p <= 100010; p += 6) criba[p] = 3;
  for (int p = 4; p <= 100010; p += 2) criba[p] = 2;
  for (int p = 7, cur = 0; p * p <= 100010; p += w[cur++ & 7])
    if (!criba[p])
      for (int j = p * p; j <= 100010; j += (p << 1))
        if (!criba[j]) criba[j] = p;
}
vector<int> primos;
void buscarPrimos() {
  crearCriba();
  for (int i = (2); i < (100010 + 1); i++)
    if (!criba[i]) primos.push_back(i);
}
void divisores(map<unsigned long long, unsigned long long> &f,
               vector<unsigned long long> &divs,
               map<unsigned long long, unsigned long long>::iterator it,
               unsigned long long n = 1) {
  if (it == f.begin()) divs.clear();
  if (it == f.end()) {
    divs.push_back(n);
    return;
  }
  unsigned long long p = it->first, k = it->second;
  ++it;
  for (int _ = (0); _ < (k + 1); _++) divisores(f, divs, it, n), n *= p;
}
void fact(unsigned long long n,
          map<unsigned long long, unsigned long long> &f) {
  for (auto p = primos.begin(); p != primos.end(); p++) {
    while (!(n % *p)) {
      f[*p]++;
      n /= *p;
    }
  }
  if (n > 1) f[n]++;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  buscarPrimos();
  unsigned long long n, k;
  cin >> n >> k;
  if (k > 1e6 || k * (k + 1) / 2 > n) {
    cout << "-1\n";
    return 0;
  }
  map<unsigned long long, unsigned long long> f;
  vector<unsigned long long> divs;
  fact(n, f);
  divisores(f, divs, f.begin());
  sort(divs.begin(), divs.end());
  int ans = 1;
  for (auto it = divs.begin(); it != divs.end(); it++) {
    if (*it * (k * (k + 1) / 2) <= n) {
      ans = *it;
    } else
      break;
  }
  unsigned long long acum = 0;
  for (int i = (1); i < (k); i++) {
    acum += ans * i;
    cout << ans * i << ' ';
  }
  cout << n - acum << '\n';
  return 0;
}

```