Problem: Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.

The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem. 

Kuroni thinks that the score distribution should satisfy the following requirements:

  * The score of each problem should be a positive integer not exceeding 10^9. 
  * A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9. 
  * The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m. 



Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.

Input

The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.

Output

If there is no solution, print a single integer -1.

Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.

Examples

Input


5 3


Output


4 5 9 13 18

Input


8 0


Output


10 11 12 13 14 15 16 17


Input


4 10


Output


-1

Note

In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution. 

  * (1, 2, 3) 
  * (1, 3, 4) 
  * (2, 4, 5) 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 10;
const int maxm = 2e5 + 10;
const int mod = 1e9 + 7;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
template <typename S, typename T>
inline bool Min(S &a, const T &b) {
  return a > b ? a = b, true : false;
}
template <typename S, typename T>
inline bool Max(S &a, const T &b) {
  return a < b ? a = b, true : false;
}
template <typename S, typename T>
inline void Adm(S &a, const T &b) {
  a = (a + b) % mod;
  if (a < 0) a += mod;
}
template <typename S, typename T>
inline void Mum(S &a, const T &b) {
  a = 1LL * a * b % mod;
}
template <typename T>
inline T Gcd(T a, T b) {
  while (b) {
    T t = b;
    b = a % b;
    a = t;
  }
  return a;
}
template <typename T>
inline int BitCnt(T x) {
  int cnt = 0;
  while (x) ++cnt, x &= x - 1;
  return cnt;
}
template <typename T>
inline bool IsPri(T x) {
  if (x < 2) return false;
  for (T i = 2; i * i <= x; ++i)
    if (x % i == 0) return false;
  return true;
}
inline long long qpow(long long a, long long n) {
  long long t = 1;
  while (n) {
    if (n & 1) t = t * a % mod;
    a = a * a % mod, n >>= 1;
  }
  return t % mod;
}
struct Bit {
  inline int lowbit(int x) { return x & (-x); }
  long long Bit[maxn];
  int N;
  bool rev = 0, MOD = 0;
  void init(int x, bool y = 0, bool z = 0) {
    N = x;
    rev = y;
    MOD = z;
    fill(Bit, Bit + N + 5, 0);
  }
  void addBitMax(int x, long long v) {
    if (rev) x = N - x + 1;
    while (x <= N) Bit[x] = max(Bit[x], v), x += lowbit(x);
  }
  void addBitSum(int x, long long v) {
    if (rev) x = N - x + 1;
    while (x <= N) {
      Bit[x] = Bit[x] + v;
      if (MOD) Bit[x] %= mod;
      x += lowbit(x);
    }
  }
  long long qBitMax(int x) {
    if (rev) x = N - x + 1;
    long long res = -linf;
    while (x) res = max(res, Bit[x]), x -= lowbit(x);
    return res;
  }
  long long qBitSum(int x) {
    if (rev) x = N - x + 1;
    long long res = 0;
    while (x) {
      res = res + Bit[x];
      if (MOD) res %= mod;
      x -= lowbit(x);
    }
    if (MOD) return res % mod;
    return res;
  }
};
void p_bit(long long x) { cout << bitset<32>(x) << ':' << x << endl; }
int a[maxn];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  if (n <= 2) {
    if (m == 0) {
      if (n == 1)
        cout << 1 << endl;
      else
        cout << "1 2" << endl;
    } else
      cout << -1 << endl;
  } else {
    a[1] = 1, a[2] = 2;
    int i;
    for (i = 3; i <= n && m; i++) {
      if ((i - 1) / 2 <= m) {
        a[i] = a[i - 1] + 1;
        m -= (i - 1) / 2;
      } else {
        a[i] = a[i - 1] + a[i - 2] - 2 * (m - 1);
        m = 0;
        ++i;
        break;
      }
    }
    if (m)
      cout << -1 << endl;
    else {
      int x = i - 1;
      for (i; i <= n; i++) a[i] = a[i - 1] + a[x] + 1;
      sort(a + 1, a + 1 + n);
      for (int i = 1; i <= n; i++) cout << a[i] << ' ';
    }
  }
  return 0;
}