Problem: Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array a with indices from [l, r) can be implemented as follows:

  1. If the segment [l, r) is already sorted in non-descending order (that is, for any i such that l ≤ i < r - 1 a[i] ≤ a[i + 1]), then end the function call; 
  2. Let <image>; 
  3. Call mergesort(a, l, mid); 
  4. Call mergesort(a, mid, r); 
  5. Merge segments [l, mid) and [mid, r), making the segment [l, r) sorted in non-descending order. The merge algorithm doesn't call any other functions. 



The array in this problem is 0-indexed, so to sort the whole array, you need to call mergesort(a, 0, n).

The number of calls of function mergesort is very important, so Ivan has decided to calculate it while sorting the array. For example, if a = {1, 2, 3, 4}, then there will be 1 call of mergesort — mergesort(0, 4), which will check that the array is sorted and then end. If a = {2, 1, 3}, then the number of calls is 3: first of all, you call mergesort(0, 3), which then sets mid = 1 and calls mergesort(0, 1) and mergesort(1, 3), which do not perform any recursive calls because segments (0, 1) and (1, 3) are sorted.

Ivan has implemented the program that counts the number of mergesort calls, but now he needs to test it. To do this, he needs to find an array a such that a is a permutation of size n (that is, the number of elements in a is n, and every integer number from [1, n] can be found in this array), and the number of mergesort calls when sorting the array is exactly k.

Help Ivan to find an array he wants!

Input

The first line contains two numbers n and k (1 ≤ n ≤ 100000, 1 ≤ k ≤ 200000) — the size of a desired permutation and the number of mergesort calls required to sort it.

Output

If a permutation of size n such that there will be exactly k calls of mergesort while sorting it doesn't exist, output  - 1. Otherwise output n integer numbers a[0], a[1], ..., a[n - 1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them.

Examples

Input

3 3


Output

2 1 3 

Input

4 1


Output

1 2 3 4 

Input

5 6


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-6;
const int PRECISION = 20;
const int MOD = 1e9 + 7;
struct node {
  long long val;
  node* next;
  node* prev;
  bool initialPos;
  node(int v) {
    val = v;
    next = nullptr;
    prev = nullptr;
    initialPos = true;
  }
};
struct group {
  long long mul, last, gcd;
  group(long long m, long long l, long long lm) {
    mul = m;
    last = l;
    gcd = lm;
  }
};
bool tryBoxes(long long mid, vector<long long> arr, long long boxes);
bool comp(pair<int, int> p1, pair<int, int> p2) {
  return p1.second < p2.second;
}
int bs(vector<int> vc, int target) {
  int low = 0, high = vc.size() - 1;
  while (high > low) {
    int mid = ((high - low + 1) / 2) + low;
    if (vc[mid] <= target)
      low = mid;
    else
      high = mid - 1;
  }
  return low;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long fastPow(long long a, long long n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  long long x = fastPow(a, n / 2);
  x = (x * x);
  return (n % 2 == 0 ? x : (x * a));
}
int char2index(char a) { return a >= 'a' ? a - 'a' : a - 'A' + 26; }
int flipCase(char a) {
  return a >= 'a' ? char2index(a - ('a' - 'A')) : char2index(a + ('a' - 'A'));
}
void prefix(vector<long long>& arr) {
  for (int i = 1; i < arr.size(); i++) arr[i] += arr[i - 1];
}
long long eval(vector<long long>& coeffs, long long x) {
  long long ans = 0;
  long long xs = 1;
  for (long long i = coeffs.size() - 1; i >= 0; i--) {
    ans += (coeffs[i] * xs);
    xs *= x;
  }
  return ans;
}
int bracketToInt(char a) {
  switch (a) {
    case '(':
      return 1;
    case ')':
      return -1;
    case '{':
      return 2;
    case '}':
      return -2;
    case '[':
      return 3;
    case ']':
      return -3;
    case '<':
      return 4;
    case '>':
      return -4;
    default:
      return -1;
  }
}
long long findOpposite(long long x, long long y, long long n) {
  long long ans = 0;
  if (x - 2 >= 1) {
    if (y - 1 >= 1) ans++;
    if (y + 1 < n) ans++;
  }
  if (x + 2 <= n) {
    if (y - 1 >= 1) ans++;
    if (y + 1 <= n) ans++;
  }
  if (y - 2 >= 1) {
    if (x + 1 <= n) ans++;
    if (x - 1 >= 1) ans++;
  }
  if (y + 2 < n) {
    if (x + 1 <= n) ans++;
    if (x - 1 >= 1) ans++;
  }
  return ans;
}
long long bs(const vector<long long>& v, long long target) {
  long long low = 0;
  long long high = v.size() - 1;
  while (low < high) {
    long long mid = ((high - low) / 2) + low;
    if (v[mid] < target)
      low = mid + 1;
    else
      high = mid;
  }
  return low;
}
bool customComp(pair<long long, long long> p1, pair<long long, long long> p2) {
  return p1.second < p2.second;
}
long long power(long long x, long long y, int p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long add(long long a, long long b) {
  return ((a % MOD) + (b % MOD)) % MOD;
}
long long mul(long long a, long long b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
long long edit_distance(const string& s, const string& t) {
  vector<vector<long long>> dp(s.size() + 1,
                               vector<long long>(t.size() + 1, 0));
  for (long long i = 0; i < s.size(); i++) {
    dp[i][0] = i;
  }
  for (long long j = 0; j < t.size(); j++) {
    dp[0][j] = j;
  }
  for (long long i = 1; i <= s.size(); i++) {
    for (long long j = 1; j <= t.size(); j++) {
      if (s[i - 1] == t[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1];
      } else {
        dp[i][j] = min(dp[i - 1][j], max(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
      }
    }
  }
  return dp[s.size()][t.size()];
}
string reverse(string a) {
  reverse(a.begin(), a.end());
  return a;
}
pair<long long, long long> palind(const vector<long long>& s, long long ind) {
  long long len = 1;
  long long start = ind, end = ind;
  while (start > 0 && end < s.size() - 1 && s[start] == s[end]) {
    start--;
    end++;
  }
  if (s[start] != s[end]) {
    start++;
    end--;
  }
  len = end - start + 1;
  pair<long long, long long> ans = make_pair(start, end);
  if (ind == s.size() - 1 || s[ind] != s[ind + 1]) return ans;
  start = ind;
  end = start + 1;
  while (start > 0 && end < s.size() - 1 && s[start] == s[end]) {
    start--;
    end++;
  }
  if (s[start] != s[end]) {
    start++;
    end--;
  }
  if (len < end - start + 1) {
    return make_pair(start, end);
  } else
    return ans;
}
string arrToString(const vector<long long>& arr) {
  string s;
  for (long long i : arr) {
    s += to_string(i) + ",";
  }
  return s;
}
bool cusSort(pair<long long, long long> p, pair<long long, long long> x) {
  if (p.first == x.first)
    return x.second < p.second;
  else
    return p.first < x.first;
}
queue<long long> q;
long long k;
void mergeCount(vector<long long>& arr, long long l, long long r) {
  if (r - l == 1) {
    arr[l] = q.front();
    q.pop();
    return;
  } else if (l >= r)
    return;
  if (k < 2) {
    for (long long i = r - 1; i >= l; i--) {
      arr[i] = q.front();
      q.pop();
    }
    return;
  }
  k -= 2;
  long long mid = (r + l) / 2;
  mergeCount(arr, l, mid);
  mergeCount(arr, mid, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n >> k;
  vector<long long> arr(n);
  for (long long i = 0; i < n; i++) q.push(n - i);
  k--;
  mergeCount(arr, 0, n);
  if (k != 0)
    cout << -1;
  else {
    for (long long i = 0; i < n; i++) {
      cout << arr[i] << ' ';
    }
  }
  return 0;
}