#include <bits/stdc++.h>
using namespace std;
int prime[150000];
int lp[1001002];
int phi[1001002];
class prime_class {
 public:
  long top;
  prime_class() {
    int i, i2, j;
    top = 0;
    lp[0] = 0;
    lp[1] = 1;
    for (i = 2; i < 1001002; i++) lp[i] = 0;
    top = 0;
    for (i = 2; i < 1001002; ++i) {
      if (lp[i] == 0) {
        lp[i] = i;
        prime[top++] = i;
      }
      for (j = 0; (j < top) && (prime[j] <= lp[i]) && (i * prime[j] < 1001002);
           ++j)
        lp[i * prime[j]] = prime[j];
    }
  }
  bool isprime(long long key) {
    if (key < 1001002)
      return (lp[key] == key) && (key >= 2);
    else {
      int i;
      for (i = 0; (i < top) && (prime[i] * prime[i] <= key); i++)
        if (key % prime[i] == 0) return false;
      return true;
    }
  }
  unordered_map<int, int> factorize(int key) {
    unordered_map<int, int> ans;
    while (lp[key] != key) {
      ans[lp[key]]++;
      key /= lp[key];
    }
    if (key > 1) ans[key]++;
    return ans;
  }
  int get_phi(int key) {
    int ans = key, last = 0;
    while (lp[key] != key) {
      if (lp[key] != last) {
        last = lp[key];
        ans -= ans / last;
      }
      key /= lp[key];
    }
    if ((key > 1) && (key != last)) ans -= ans / key;
    return ans;
  }
  void calc_all_phi(int n) {
    int i, j, k;
    for (int i = 1; i < n; i++) phi[i] = i;
    for (int i = 2; i < n; i++) {
      if (phi[i] == i) {
        for (int j = i; j < n; j += i) {
          phi[j] /= i;
          phi[j] *= i - 1;
        }
      }
    }
  }
  vector<pair<long long, long long>> factorize_full(long long key) {
    vector<pair<long long, long long>> ans;
    long i, ct, sq = sqrt(key) + 10;
    for (i = 0; (i < top) && (prime[i] <= sq); i++)
      if (key % prime[i] == 0) {
        ct = 0;
        while (key % prime[i] == 0) {
          ct++;
          key /= prime[i];
        }
        ans.push_back({prime[i], ct});
      }
    if (key > 1) {
      ans.push_back({key, 1});
    }
    return ans;
  }
  void generate_divisors(int step, int v, vector<pair<int, int>>& fp,
                         vector<int>& ans) {
    if (step < fp.size()) {
      generate_divisors(step + 1, v, fp, ans);
      for (int i = 1; i <= fp[step].second; i++) {
        v *= fp[step].first;
        generate_divisors(step + 1, v, fp, ans);
      }
    } else
      ans.push_back(v);
  }
  void generate_divisors_full(long long step, long long v,
                              vector<pair<long long, long long>>& fp,
                              vector<long long>& ans) {
    if (step < fp.size()) {
      generate_divisors_full(step + 1, v, fp, ans);
      for (int i = 1; i <= fp[step].second; i++) {
        v *= fp[step].first;
        generate_divisors_full(step + 1, v, fp, ans);
      }
    } else
      ans.push_back(v);
  }
  vector<int> get_divisors(int key) {
    unordered_map<int, int> f = factorize(key);
    int n = f.size();
    vector<pair<int, int>> fp;
    for (auto p : f) fp.push_back(p);
    vector<int> ans;
    generate_divisors(0, 1, fp, ans);
    return ans;
  }
  vector<long long> get_divisors_full(long long key) {
    vector<pair<long long, long long>> f = factorize_full(key);
    int n = f.size();
    vector<pair<long long, long long>> fp;
    for (auto p : f) fp.push_back(p);
    vector<long long> ans;
    generate_divisors_full(0, 1, fp, ans);
    return ans;
  }
  long long get_divisors_count(long long key) {
    vector<pair<long long, long long>> f = factorize_full(key);
    long long ans = 1;
    for (auto p : f) ans *= (p.second + 1);
    return ans;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k, i, j, ans, s, v;
  long double vv;
  prime_class pr;
  cin >> n >> k;
  vv = 1.0 * (1.0 + k) * k / 2.0;
  if ((n < k) || (n * 1.0 < vv))
    cout << "-1\n";
  else {
    vector<long long> d = pr.get_divisors_full(n);
    ans = 1;
    v = (1 + k) * k / 2;
    sort(d.rbegin(), d.rend());
    for (auto p : d) {
      if (n / p >= v) {
        s = n;
        for (i = 1; i <= k - 1; i++) {
          cout << p * i << " ";
          s -= p * i;
        }
        cout << s << "\n";
        break;
      }
    }
  }
  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;
int prime[150000];
int lp[1001002];
int phi[1001002];
class prime_class {
 public:
  long top;
  prime_class() {
    int i, i2, j;
    top = 0;
    lp[0] = 0;
    lp[1] = 1;
    for (i = 2; i < 1001002; i++) lp[i] = 0;
    top = 0;
    for (i = 2; i < 1001002; ++i) {
      if (lp[i] == 0) {
        lp[i] = i;
        prime[top++] = i;
      }
      for (j = 0; (j < top) && (prime[j] <= lp[i]) && (i * prime[j] < 1001002);
           ++j)
        lp[i * prime[j]] = prime[j];
    }
  }
  bool isprime(long long key) {
    if (key < 1001002)
      return (lp[key] == key) && (key >= 2);
    else {
      int i;
      for (i = 0; (i < top) && (prime[i] * prime[i] <= key); i++)
        if (key % prime[i] == 0) return false;
      return true;
    }
  }
  unordered_map<int, int> factorize(int key) {
    unordered_map<int, int> ans;
    while (lp[key] != key) {
      ans[lp[key]]++;
      key /= lp[key];
    }
    if (key > 1) ans[key]++;
    return ans;
  }
  int get_phi(int key) {
    int ans = key, last = 0;
    while (lp[key] != key) {
      if (lp[key] != last) {
        last = lp[key];
        ans -= ans / last;
      }
      key /= lp[key];
    }
    if ((key > 1) && (key != last)) ans -= ans / key;
    return ans;
  }
  void calc_all_phi(int n) {
    int i, j, k;
    for (int i = 1; i < n; i++) phi[i] = i;
    for (int i = 2; i < n; i++) {
      if (phi[i] == i) {
        for (int j = i; j < n; j += i) {
          phi[j] /= i;
          phi[j] *= i - 1;
        }
      }
    }
  }
  vector<pair<long long, long long>> factorize_full(long long key) {
    vector<pair<long long, long long>> ans;
    long i, ct, sq = sqrt(key) + 10;
    for (i = 0; (i < top) && (prime[i] <= sq); i++)
      if (key % prime[i] == 0) {
        ct = 0;
        while (key % prime[i] == 0) {
          ct++;
          key /= prime[i];
        }
        ans.push_back({prime[i], ct});
      }
    if (key > 1) {
      ans.push_back({key, 1});
    }
    return ans;
  }
  void generate_divisors(int step, int v, vector<pair<int, int>>& fp,
                         vector<int>& ans) {
    if (step < fp.size()) {
      generate_divisors(step + 1, v, fp, ans);
      for (int i = 1; i <= fp[step].second; i++) {
        v *= fp[step].first;
        generate_divisors(step + 1, v, fp, ans);
      }
    } else
      ans.push_back(v);
  }
  void generate_divisors_full(long long step, long long v,
                              vector<pair<long long, long long>>& fp,
                              vector<long long>& ans) {
    if (step < fp.size()) {
      generate_divisors_full(step + 1, v, fp, ans);
      for (int i = 1; i <= fp[step].second; i++) {
        v *= fp[step].first;
        generate_divisors_full(step + 1, v, fp, ans);
      }
    } else
      ans.push_back(v);
  }
  vector<int> get_divisors(int key) {
    unordered_map<int, int> f = factorize(key);
    int n = f.size();
    vector<pair<int, int>> fp;
    for (auto p : f) fp.push_back(p);
    vector<int> ans;
    generate_divisors(0, 1, fp, ans);
    return ans;
  }
  vector<long long> get_divisors_full(long long key) {
    vector<pair<long long, long long>> f = factorize_full(key);
    int n = f.size();
    vector<pair<long long, long long>> fp;
    for (auto p : f) fp.push_back(p);
    vector<long long> ans;
    generate_divisors_full(0, 1, fp, ans);
    return ans;
  }
  long long get_divisors_count(long long key) {
    vector<pair<long long, long long>> f = factorize_full(key);
    long long ans = 1;
    for (auto p : f) ans *= (p.second + 1);
    return ans;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k, i, j, ans, s, v;
  long double vv;
  prime_class pr;
  cin >> n >> k;
  vv = 1.0 * (1.0 + k) * k / 2.0;
  if ((n < k) || (n * 1.0 < vv))
    cout << "-1\n";
  else {
    vector<long long> d = pr.get_divisors_full(n);
    ans = 1;
    v = (1 + k) * k / 2;
    sort(d.rbegin(), d.rend());
    for (auto p : d) {
      if (n / p >= v) {
        s = n;
        for (i = 1; i <= k - 1; i++) {
          cout << p * i << " ";
          s -= p * i;
        }
        cout << s << "\n";
        break;
      }
    }
  }
  return 0;
}

```