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>
const double eps = (1e-9);
using namespace std;
int dcmp(long double a, long double b) {
  return fabsl(a - b) <= eps ? 0 : (a > b) ? 1 : -1;
}
int getBit(long long num, int idx) { return ((num >> idx) & 1ll) == 1; }
int setBit1(long long num, long long idx) { return num | (1ll << idx); }
long long setBit0(long long num, int idx) { return num & ~(1ll << idx); }
long long flipBit(long long num, int idx) { return num ^ (1ll << idx); }
void run() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int countNumBit1(int mask) {
  int ret = 0;
  while (mask) {
    mask &= (mask - 1);
    ++ret;
  }
  return ret;
}
const int N = 5000 + 9;
long long n, m, ans[N];
int cnt[N];
void print() {
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
}
int main() {
  run();
  cin >> n >> m;
  ans[1] = 1;
  ans[2] = 2;
  for (int i = 3; i <= n; i++) {
    cnt[i] = (i - 1) / 2;
  }
  int need = m, ls = 2;
  for (int i = 3; i <= n && need > 0; i++) {
    if (need >= cnt[i]) {
      ans[i] = ans[i - 1] + 1;
      need -= cnt[i];
      ls = i;
    } else {
      if ((i - 1) % 2 == 0) {
        int diff = cnt[i] - need;
        ans[i] = (ans[i - 1] + 1) + (diff - 1) * 2 + 1;
        ls = i;
        need = 0;
        break;
      } else {
        int diff = cnt[i] - need;
        ans[i] = (ans[i - 1] + 1) + (diff)*2;
        ls = i;
        need = 0;
        break;
      }
    }
  }
  if (need) {
    cout << "-1\n";
    return 0;
  }
  if (ls == n) {
    print();
    return 0;
  }
  for (int i = ls + 1; i <= n; i++) {
    ans[i] = 100000000 + i * (10001);
  }
  print();
}