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;
const signed long long Infinity = 1000000001;
const long double Epsilon = 1e-9;
template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  return os << "(" << p.first << "," << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& V) {
  os << "[";
  for (__typeof(V.begin()) i = V.begin(); i != V.end(); ++i)
    os << *i << (i - V.begin() + 1 == ((int)V.size()) ? "" : ",");
  os << "]\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& S) {
  os << "(";
  for (__typeof(S.begin()) i = S.begin(); i != S.end(); ++i)
    os << *i << (*i == *S.rbegin() ? "" : ",");
  os << ")\n";
  return os;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, const map<T, U>& M) {
  os << "{";
  for (__typeof(M.begin()) i = M.begin(); i != M.end(); ++i)
    os << *i << (i->first == M.rbegin()->first ? "" : ",");
  os << "}\n";
  return os;
}
int N, K;
void read_data() { scanf("%d %d", &N, &K); }
void solve() {
  if (N == K) {
    printf("-1\n");
  } else {
    for (int(i) = (1); (i) <= (N - K - 1); (i)++) printf("%d ", i + 1);
    printf("1 ");
    for (int(i) = (N - K + 1); (i) <= (N); (i)++) printf("%d ", i);
  }
}
int main() {
  read_data();
  solve();
  return 0;
}