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;
using lli = long long int;
using vb = vector<bool>;
using vc = vector<char>;
using vd = vector<double>;
using vi = vector<int>;
using vvi = vector<vi>;
using vlli = vector<lli>;
using vvlli = vector<vlli>;
using pi = pair<int, int>;
using plli = pair<lli, lli>;
void readc(char &q) { scanf("%c", &q); }
char readc() {
  char q;
  scanf("%c", &q);
  return q;
}
void readint(int &q) { scanf("%d", &q); }
int readint() {
  int q;
  scanf("%d", &q);
  return q;
}
void readlong(lli &q) { scanf("%lld", &q); }
lli readlong() {
  lli q;
  scanf("%lld", &q);
  return q;
}
void readdbl(double &q) { scanf("%lf", &q); }
double readdbl() {
  double q;
  scanf("%lf", &q);
  return q;
}
void printint(int q) { printf("%d", q); }
void printlong(lli q) { printf("%lld", q); }
inline void printdbl(double q, int p = 6) { printf("%.*lf", p, q); }
void readln() { scanf("\n"); }
void println() { printf("\n"); }
void printsp() { printf(" "); }
void print(int q) { printint(q); }
void print(long q) { printlong((lli)q); }
void print(lli q) { printlong(q); }
void print(double q) { printdbl(q); }
inline void print(char c) { printf("%c", c); }
void print(pi q) {
  print(q.first);
  printsp();
  print(q.second);
}
void print(vi q) {
  for (int i = 0; i < q.size(); ++i) {
    print(q[i]);
    printsp();
  }
}
void print(vlli q) {
  for (int i = 0; i < q.size(); ++i) {
    print(q[i]);
    printsp();
  }
}
void print(vd q) {
  for (int i = 0; i < q.size(); ++i) {
    print(q[i]);
    printsp();
  }
}
template <typename T>
void println(T q) {
  print(q);
  println();
}
template <typename T>
T maxim(T first, T second) {
  return (first > second) ? first : second;
}
template <typename T>
T maxim(T first, T second, T third) {
  return mndlim(maxim(first, second), third);
}
template <typename T>
T minim(T first, T second) {
  return (first < second) ? first : second;
}
template <typename T>
T minim(T first, T second, T third) {
  return minim(minim(first, second), third);
}
template <typename T>
T middle(T first, T second, T third) {
  return first + second + third - maxim(first, second, third) -
         minim(first, second, third);
}
template <typename T>
T abs(T arg) {
  return arg < 0 ? -arg : arg;
}
const int modulo = 1000000007;
int a[5005];
vlli ans;
lli create_stint(lli base, int size) {
  for (int i = 0; i < size; ++i) {
    ans.push_back(base * (i + 1));
  }
  return *ans.rbegin() * 2;
}
int main() {
  int n = readint();
  int m = readint();
  a[3] = 1;
  for (int i = 4; i <= n + 1; ++i) {
    a[i] = a[i - 1] + (i - 1) / 2;
  }
  int cnt = n;
  lli base = 1ll;
  if (m > a[n]) {
    println(-1);
    return 0;
  }
  int cur = 2;
  while (a[cur] <= m) {
    cur++;
  }
  cur--;
  for (int i = 0; i < cur; ++i) ans.push_back(i + 1);
  cnt -= cur;
  m -= a[cur];
  int last = *ans.rbegin();
  if (m > 0) {
    if (cnt > 0) {
      ans.push_back(2 * last - 2 * m + 1);
      cnt--;
      m = 0;
    } else {
      println(-1);
      return 0;
    }
  }
  last = *ans.rbegin();
  for (int i = 0; i < cnt; ++i) ans.push_back(100000 * (i + 1) + i);
  for (int i = 0; i < n; ++i) cout << ans[i] << " ";
  println();
  return 0;
}