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 MOD = 1000000007;
long long POW(long long a, long long b, long long MMM = MOD) {
  long long ret = 1;
  for (; b; b >>= 1, a = (a * a) % MMM)
    if (b & 1) ret = (ret * a) % MMM;
  return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) {
  if (a == 0 || b == 0) return a + b;
  return a * (b / gcd(a, b));
}
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
int ddx[] = {-1, -2, 1, -2, 2, -1, 2, 1}, ddy[] = {-2, -1, -2, 1, -1, 2, 1, 2};
int a[5005];
int main() {
  for (int(i) = (1); (i) <= (5000); (i) += (1)) a[i] = i - 1 >> 1;
  int n, m;
  scanf("%d%d", &n, &m);
  if (accumulate(a, a + n + 1, 0) < m) return !printf("-1");
  if (m == 0) {
    for (int(i) = (1000000001 - n); (i) <= (1000000000); (i) += (1))
      printf("%d ", i);
    return 0;
  }
  vector<int> ans = {1, 2};
  int now = 0;
  for (int(i) = (3); (i) <= (n); (i) += (1)) {
    if (now < m && m <= now + a[i]) {
      for (int(j) = (i + 1); (j) <= (n); (j) += (1)) {
        int t = n - j;
        ans.push_back(1000000000 - 10005 * t);
      }
      int tt = i, rem = m - now;
      rem = a[i] - rem;
      for (int(i) = (1); (i) <= (rem); (i) += (1)) tt += 2;
      ans.push_back(tt);
      break;
    }
    now += a[i];
    ans.push_back(i);
  }
  sort(ans.begin(), ans.end());
  for (int x : ans) printf("%d ", x);
}