#include <bits/stdc++.h>
using namespace std;
const long long N = 100001;
long long A[N];
bool vis[N];
long long ex[20];
int main() {
  ex[0] = 1;
  for (long long i = 1; i < 20; i++) {
    ex[i] = ex[i - 1] * 2;
  }
  for (long long i = 1; i < N; i++) {
    for (long long j = 0; j < 20; j++) {
      if ((i >> j) & 1 == 1) {
        A[i] = ex[j];
        break;
      }
    }
  }
  long long sum, lim;
  cin >> sum >> lim;
  vector<long long> v;
  map<long long, vector<long long> > puede;
  for (long long i = 1; i <= lim; i++) {
    v.push_back(A[i]);
    puede[A[i]].push_back(i);
  }
  sort(v.begin(), v.end());
  vector<long long> rep;
  while (sum > 0) {
    vector<long long>::iterator upp;
    upp = upper_bound(v.begin(), v.end(), sum);
    long long pos = upp - v.begin() - 1;
    sum -= v[pos];
    vector<long long> aux;
    aux = puede[v[pos]];
    for (long long i = 0; i < aux.size(); i++) {
      if (!vis[aux[i]]) {
        vis[aux[i]] = true;
        rep.push_back(aux[i]);
        break;
      }
    }
    v.erase(v.begin() + pos);
    long long sz = v.size();
    if (sz == 0) {
      break;
    }
  }
  if (sum == 0) {
    long long sz = rep.size();
    cout << sz << endl;
    for (long long i = 0; i < sz; i++) {
      cout << rep[i] << " ";
    }
    cout << endl;
  } else {
    cout << -1 << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite set of Picks.

Fortunately, Picks remembers something about his set S:

  * its elements were distinct integers from 1 to limit; 
  * the value of <image> was equal to sum; here lowbit(x) equals 2k where k is the position of the first one in the binary representation of x. For example, lowbit(100102) = 102, lowbit(100012) = 12, lowbit(100002) = 100002 (binary representation). 



Can you help Picks and find any set S, that satisfies all the above conditions?

Input

The first line contains two integers: sum, limit (1 ≤ sum, limit ≤ 105).

Output

In the first line print an integer n (1 ≤ n ≤ 105), denoting the size of S. Then print the elements of set S in any order. If there are multiple answers, print any of them.

If it's impossible to find a suitable set, print -1.

Examples

Input

5 5


Output

2
4 5


Input

4 3


Output

3
2 3 1


Input

5 1


Output

-1

Note

In sample test 1: lowbit(4) = 4, lowbit(5) = 1, 4 + 1 = 5.

In sample test 2: lowbit(1) = 1, lowbit(2) = 2, lowbit(3) = 1, 1 + 2 + 1 = 4.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long N = 100001;
long long A[N];
bool vis[N];
long long ex[20];
int main() {
  ex[0] = 1;
  for (long long i = 1; i < 20; i++) {
    ex[i] = ex[i - 1] * 2;
  }
  for (long long i = 1; i < N; i++) {
    for (long long j = 0; j < 20; j++) {
      if ((i >> j) & 1 == 1) {
        A[i] = ex[j];
        break;
      }
    }
  }
  long long sum, lim;
  cin >> sum >> lim;
  vector<long long> v;
  map<long long, vector<long long> > puede;
  for (long long i = 1; i <= lim; i++) {
    v.push_back(A[i]);
    puede[A[i]].push_back(i);
  }
  sort(v.begin(), v.end());
  vector<long long> rep;
  while (sum > 0) {
    vector<long long>::iterator upp;
    upp = upper_bound(v.begin(), v.end(), sum);
    long long pos = upp - v.begin() - 1;
    sum -= v[pos];
    vector<long long> aux;
    aux = puede[v[pos]];
    for (long long i = 0; i < aux.size(); i++) {
      if (!vis[aux[i]]) {
        vis[aux[i]] = true;
        rep.push_back(aux[i]);
        break;
      }
    }
    v.erase(v.begin() + pos);
    long long sz = v.size();
    if (sz == 0) {
      break;
    }
  }
  if (sum == 0) {
    long long sz = rep.size();
    cout << sz << endl;
    for (long long i = 0; i < sz; i++) {
      cout << rep[i] << " ";
    }
    cout << endl;
  } else {
    cout << -1 << endl;
  }
  return 0;
}

```