#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 4e18;
const long long MAX = 1e5;
const long double PI = 3.14159265358979323846;
bool IN_RANGE(pair<long long, long long> v, long long n, long long m,
              long long d) {
  return v.first >= d && v.first <= n && v.second >= d && v.second <= m;
}
bool IN_R(long long x, long long n, long long d) { return x >= d && x <= n; }
long long MA = -INF, MI = INF;
long long C2I(char x, char base) { return x - base; }
void PAIROUT(pair<long long, long long> x) {
  cout << x.first << " " << x.second;
}
long long PO(long long x, long long y) {
  if (y == 0) return 1;
  long long ans = PO(x, y / 2);
  ans *= ans, ans %= MOD;
  if (y % 2 == 1) ans *= x, ans %= MOD;
  return ans % MOD;
}
void RE(long long x) {
  x %= MOD;
  if (x < 0) x += MOD;
}
long long up(long long n, long long k) {
  if (n % k == 0)
    return n / k;
  else
    return n / k + 1;
}
set<long long> di;
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0),
      cout << fixed << setprecision(10);
  long long n, k;
  cin >> n >> k;
  if (n == 1 && k == 1) return cout << 1, 0;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) di.insert(i), di.insert(n / i);
  }
  vector<long long> temp;
  for (long long u : di) temp.push_back(u);
  reverse(temp.begin(), temp.end());
  long long gcd;
  bool can = false;
  for (long long u : temp) {
    gcd = n / u;
    if (k < gcd && k + 1 <= (2 * gcd) / k) {
      can = 1;
      break;
    }
  }
  if (!can) return cout << -1, 0;
  long long e = gcd;
  for (long long i = 1; i <= k - 1; i++) cout << i * (n / gcd) << " ", e -= i;
  cout << e * (n / gcd);
  return 0;
}

### Prompt

Please formulate 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;
const long long MOD = 1e9 + 7;
const long long INF = 4e18;
const long long MAX = 1e5;
const long double PI = 3.14159265358979323846;
bool IN_RANGE(pair<long long, long long> v, long long n, long long m,
              long long d) {
  return v.first >= d && v.first <= n && v.second >= d && v.second <= m;
}
bool IN_R(long long x, long long n, long long d) { return x >= d && x <= n; }
long long MA = -INF, MI = INF;
long long C2I(char x, char base) { return x - base; }
void PAIROUT(pair<long long, long long> x) {
  cout << x.first << " " << x.second;
}
long long PO(long long x, long long y) {
  if (y == 0) return 1;
  long long ans = PO(x, y / 2);
  ans *= ans, ans %= MOD;
  if (y % 2 == 1) ans *= x, ans %= MOD;
  return ans % MOD;
}
void RE(long long x) {
  x %= MOD;
  if (x < 0) x += MOD;
}
long long up(long long n, long long k) {
  if (n % k == 0)
    return n / k;
  else
    return n / k + 1;
}
set<long long> di;
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0),
      cout << fixed << setprecision(10);
  long long n, k;
  cin >> n >> k;
  if (n == 1 && k == 1) return cout << 1, 0;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) di.insert(i), di.insert(n / i);
  }
  vector<long long> temp;
  for (long long u : di) temp.push_back(u);
  reverse(temp.begin(), temp.end());
  long long gcd;
  bool can = false;
  for (long long u : temp) {
    gcd = n / u;
    if (k < gcd && k + 1 <= (2 * gcd) / k) {
      can = 1;
      break;
    }
  }
  if (!can) return cout << -1, 0;
  long long e = gcd;
  for (long long i = 1; i <= k - 1; i++) cout << i * (n / gcd) << " ", e -= i;
  cout << e * (n / gcd);
  return 0;
}

```