Problem: Levko loves permutations very much. A permutation of length n is a sequence of distinct positive integers, each is at most n.

Let’s assume that value gcd(a, b) shows the greatest common divisor of numbers a and b. Levko assumes that element pi of permutation p1, p2, ... , pn is good if gcd(i, pi) > 1. Levko considers a permutation beautiful, if it has exactly k good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them.

Input

The single line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ n).

Output

In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.

If there are multiple suitable permutations, you are allowed to print any of them.

Examples

Input

4 2


Output

2 4 3 1

Input

1 1


Output

-1

Note

In the first sample elements 4 and 3 are good because gcd(2, 4) = 2 > 1 and gcd(3, 3) = 3 > 1. Elements 2 and 1 are not good because gcd(1, 2) = 1 and gcd(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.

The second sample has no beautiful permutations.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
const long double pi = 3.14159265359;
const long long mod = 1000000007;
const long long modc = 998244353;
const long long inf = LLONG_MAX;
const long long neg_inf = LLONG_MIN;
long long inline power(long long a, long long b, long long p) {
  a %= p;
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return ans;
}
long long mod_inv(long long n, long long p) { return power(n, p - 2, p); }
bool inline isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
template <typename... T>
void W(T &&...args) {
  ((cout << fixed << setprecision(17) << args << " "), ...);
  cout << endl;
}
template <typename... T>
void R(T &&...args) {
  ((cin >> args), ...);
}
long long ston(string s) {
  long long a = 0;
  reverse(s.begin(), s.end());
  long long pow = 1;
  for (long long i = 0; i < s.size(); i++) {
    a += pow * (s[i] - '0');
    pow *= 10;
  }
  return a;
}
template <class T>
string to_binary(T n) {
  string binaryNum;
  long long i = 0;
  while (n > 0) {
    binaryNum.push_back((n % 2) + '0');
    n = n / 2;
    i++;
  }
  if ((long long)binaryNum.size() == 0) binaryNum.push_back('0');
  reverse(binaryNum.begin(), binaryNum.end());
  return binaryNum;
}
long long bton(string &s, long long base) {
  long long ans = 0;
  long long pow = 1;
  reverse(s.begin(), s.end());
  for (auto x : s) {
    if (x >= 'a' && x <= 'z')
      ans += (x - 'a' + 10) * pow;
    else
      ans += (x - '0') * pow;
    pow *= base;
  }
  return ans;
}
template <class T>
void wv(T &v) {
  for (auto x : v) cout << x << " ";
  cout << endl;
}
template <class T>
void rv(T &v) {
  for (long long i = 0; i < (long long)v.size(); i++) cin >> v[i];
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
const long long M = 2e5;
void solve() {
  long long i, j, sum = 0, prod = 1, maxl = neg_inf, minl = inf;
  long long n, k;
  cin >> n >> k;
  if (k == n) {
    cout << -1 << endl;
    return;
  }
  vector<long long> ans(n + 1);
  for (i = n; i > n - k; i--) ans[i] = i;
  ans[1] = n - k;
  for (i = 2; i <= n - k; i++) ans[i] = i - 1;
  for (i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) {
    solve();
  }
}