#include <bits/stdc++.h>
using namespace std;
long long int MOD = 1000000007;
long long int inf = 1e15;
void scan(int &x);
long long int powermod(long long int _a, long long int _b, long long int _m) {
  long long int _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a) % _m;
    _b /= 2;
    _a = (_a * _a) % _m;
  }
  return _r;
}
long long int string_to_number(string s) {
  long long int x = 0;
  stringstream convert(s);
  convert >> x;
  return x;
}
long long int add(long long int a, long long int b) {
  long long int x = (a + b) % MOD;
  return x;
}
long long int mul(long long int a, long long int b) {
  long long int x = (a * b) % MOD;
  return x;
}
long long int sub(long long int a, long long int b) {
  long long int x = (a - b + MOD) % MOD;
  return x;
}
long long int divi(long long int a, long long int b) {
  long long int x = a;
  long long int y = powermod(b, MOD - 2, MOD);
  long long int res = (x * y) % MOD;
  return res;
}
int main(void) {
  long long int n, k;
  cin >> n >> k;
  if (k > 1e6) {
    cout << -1;
    return 0;
  }
  long long int lsq = sqrt(n), sumLim = (k * (k + 1LL)) / 2LL;
  long long int mmx = -1, mul;
  for (mul = 1; mul <= lsq; mul++)
    if (n % mul == 0 && n / mul >= sumLim) {
      mmx = max(mmx, mul);
      if (mul >= sumLim) mmx = max(mmx, n / mul);
    }
  if (mmx == -1) {
    cout << -1;
    return 0;
  }
  mul = mmx;
  long long int rsum = n / mul;
  long long int lastElement = rsum - ((k * (k - 1LL)) / 2LL);
  for (int i = 1; i < k; i++) printf("%lld ", 1LL * i * mul);
  printf("%lld", lastElement * mul);
  return 0;
}
void scan(int &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}

### 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;
long long int MOD = 1000000007;
long long int inf = 1e15;
void scan(int &x);
long long int powermod(long long int _a, long long int _b, long long int _m) {
  long long int _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a) % _m;
    _b /= 2;
    _a = (_a * _a) % _m;
  }
  return _r;
}
long long int string_to_number(string s) {
  long long int x = 0;
  stringstream convert(s);
  convert >> x;
  return x;
}
long long int add(long long int a, long long int b) {
  long long int x = (a + b) % MOD;
  return x;
}
long long int mul(long long int a, long long int b) {
  long long int x = (a * b) % MOD;
  return x;
}
long long int sub(long long int a, long long int b) {
  long long int x = (a - b + MOD) % MOD;
  return x;
}
long long int divi(long long int a, long long int b) {
  long long int x = a;
  long long int y = powermod(b, MOD - 2, MOD);
  long long int res = (x * y) % MOD;
  return res;
}
int main(void) {
  long long int n, k;
  cin >> n >> k;
  if (k > 1e6) {
    cout << -1;
    return 0;
  }
  long long int lsq = sqrt(n), sumLim = (k * (k + 1LL)) / 2LL;
  long long int mmx = -1, mul;
  for (mul = 1; mul <= lsq; mul++)
    if (n % mul == 0 && n / mul >= sumLim) {
      mmx = max(mmx, mul);
      if (mul >= sumLim) mmx = max(mmx, n / mul);
    }
  if (mmx == -1) {
    cout << -1;
    return 0;
  }
  mul = mmx;
  long long int rsum = n / mul;
  long long int lastElement = rsum - ((k * (k - 1LL)) / 2LL);
  for (int i = 1; i < k; i++) printf("%lld ", 1LL * i * mul);
  printf("%lld", lastElement * mul);
  return 0;
}
void scan(int &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}

```