#include <bits/stdc++.h>
using namespace std;
using vint = vector<long long>;
using pint = pair<long long, long long>;
using vpint = vector<pint>;
template <typename A, typename B>
inline void chmin(A& a, B b) {
  if (a > b) a = b;
}
template <typename A, typename B>
inline void chmax(A& a, B b) {
  if (a < b) a = b;
}
template <class A, class B>
ostream& operator<<(ostream& ost, const pair<A, B>& p) {
  ost << "{" << p.first << "," << p.second << "}";
  return ost;
}
template <class T>
ostream& operator<<(ostream& ost, const vector<T>& v) {
  ost << "{";
  for (long long i = 0; i < v.size(); i++) {
    if (i) ost << ",";
    ost << v[i];
  }
  ost << "}";
  return ost;
}
inline long long topbit(unsigned long long x) {
  return x ? 63 - __builtin_clzll(x) : -1;
}
inline long long popcount(unsigned long long x) {
  return __builtin_popcountll(x);
}
inline long long parity(unsigned long long x) { return __builtin_parity(x); }
signed main() {
  long long N, K;
  cin >> N >> K;
  if (K > 1000000 || K * (K + 1) / 2 > N) {
    cout << -1 << endl;
    return 0;
  }
  long long s = K * (K + 1) / 2;
  long long G = 0;
  for (long long g = 1; g * g <= N; g++) {
    if (N % g) continue;
    if (N / s >= g) chmax(G, g);
    if (N / s >= N / g) chmax(G, N / g);
  }
  vint ans(K);
  iota((ans).begin(), (ans).end(), 1);
  ans[K - 1] += N / G - s;
  for (long long i = 0; i < (K); i++) ans[i] *= G;
  for (long long i = 0; i < (K); i++) {
    if (i) cout << " ";
    cout << ans[i];
  }
  cout << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
using vint = vector<long long>;
using pint = pair<long long, long long>;
using vpint = vector<pint>;
template <typename A, typename B>
inline void chmin(A& a, B b) {
  if (a > b) a = b;
}
template <typename A, typename B>
inline void chmax(A& a, B b) {
  if (a < b) a = b;
}
template <class A, class B>
ostream& operator<<(ostream& ost, const pair<A, B>& p) {
  ost << "{" << p.first << "," << p.second << "}";
  return ost;
}
template <class T>
ostream& operator<<(ostream& ost, const vector<T>& v) {
  ost << "{";
  for (long long i = 0; i < v.size(); i++) {
    if (i) ost << ",";
    ost << v[i];
  }
  ost << "}";
  return ost;
}
inline long long topbit(unsigned long long x) {
  return x ? 63 - __builtin_clzll(x) : -1;
}
inline long long popcount(unsigned long long x) {
  return __builtin_popcountll(x);
}
inline long long parity(unsigned long long x) { return __builtin_parity(x); }
signed main() {
  long long N, K;
  cin >> N >> K;
  if (K > 1000000 || K * (K + 1) / 2 > N) {
    cout << -1 << endl;
    return 0;
  }
  long long s = K * (K + 1) / 2;
  long long G = 0;
  for (long long g = 1; g * g <= N; g++) {
    if (N % g) continue;
    if (N / s >= g) chmax(G, g);
    if (N / s >= N / g) chmax(G, N / g);
  }
  vint ans(K);
  iota((ans).begin(), (ans).end(), 1);
  ans[K - 1] += N / G - s;
  for (long long i = 0; i < (K); i++) ans[i] *= G;
  for (long long i = 0; i < (K); i++) {
    if (i) cout << " ";
    cout << ans[i];
  }
  cout << endl;
  return 0;
}

```