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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
string toBin(long long a) {
  string res = "";
  while (a) {
    res += char((a & 1) + '0');
    a >>= 1;
  }
  reverse(res.begin(), res.end());
  return res;
}
const int mod = 1e9 + 7;
long long expo(long long base, long long pow) {
  long long ans = 1;
  while (pow != 0) {
    if (pow & 1 == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base *= base;
    base %= mod;
    pow /= 2;
  }
  return ans;
}
long long inv(long long x) { return expo(x, mod - 2); }
bool isPal(string ss) {
  int len = ss.length();
  for (int i = 0; i < len / 2; i++) {
    int comp = len - i - 1;
    if (ss[i] != ss[comp]) return false;
  }
  return true;
}
double pi = 3.141592653589793238462643;
double error = 0.0000001;
const int M = 100001;
int arr[M + 21];
int main() {
  int n, k;
  scanf("%d", &(n));
  scanf("%d", &(k));
  int ma = n - 1;
  if (k > ma) {
    cout << "-1";
    return 0;
  }
  int left = n - k;
  if (left & 1) {
    for (int i = n; i > n - k; i--) {
      arr[i] = i;
    }
    arr[1] = 1;
    for (int i = 2; i <= left; i += 2) {
      arr[i] = i + 1;
      arr[i + 1] = i;
    }
  } else {
    for (int i = n; i > n - k; i--) {
      arr[i] = i;
    }
    for (int i = 1; i <= left; i += 2) {
      arr[i] = i + 1;
      arr[i + 1] = i;
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", arr[i]);
  }
}
