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.
#include <bits/stdc++.h>
using namespace std;
struct pair {
  int a;
  int b;
};
bool isPrime(int x) {
  if (x == 1) {
    return false;
  }
  long long a = (long long)x;
  for (int i = 2; i <= (long long)sqrt((double)a); i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
int gcd(int a, int b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long binToInt(string s) {
  long long x = 0;
  int j = 0;
  for (int i = s.length() - 1; i >= 0; i--) {
    x += (long long)pow(2, (double)j) * (s[i] - '0');
    j++;
  }
  return x;
}
long long factorial(long long x) {
  long long sum = 1;
  for (long long i = 1; i <= x; i++) {
    sum *= i;
  }
  return sum;
}
long long calc(long long a, long long b) { return (a * b) / gcd(a, b); }
bool isPalindrom(int x) {
  int t = x;
  int j = 1;
  while (t != 0) {
    j *= 10;
    t /= 10;
  }
  j /= 10;
  int q = x;
  t = 0;
  while (q != 0) {
    t += q % 10 * j;
    j /= 10;
    q /= 10;
  }
  return t == x;
}
int a[100005];
int main() {
  int n, k;
  a[n];
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    a[i] = 0;
  }
  if (n == 1 && k == 0) {
    cout << 1;
    return 0;
  }
  if (n == 1 || k == n) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= k; i++) {
    a[i] = i + 1;
  }
  if (n - k == 1) {
    a[0] = 1;
    for (int i = 0; i < n; i++) {
      cout << a[i] << " ";
    }
    return 0;
  }
  int j = k + 2;
  for (int i = 0; i < n - 1; i++) {
    if (a[i] == 0) {
      a[i] = j;
      j++;
    }
  }
  a[n - 1] = 1;
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
}
