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;
inline double Time() { return (clock() * 1.0) / CLOCKS_PER_SEC; }
inline void mem(int a[], int x, int l, int r) {
  for (int i = l; i <= r; i++) a[i] = x;
}
inline void mem_arr(int a[], int b[], int l, int r) {
  for (int i = l; i <= r; i++) a[i] = b[i];
}
const int N = 100500, inf = 1e9 * 2;
const long long MOD = 1e9 + 7ll, INF = 1e18;
const int dx[] = {1, -1, 0, 0, -1, 1, -1, 1};
const int dy[] = {0, 0, 1, -1, -1, 1, 1, -1};
int a[N];
int main() {
  unsigned int FOR;
  asm("rdtsc" : "=A"(FOR));
  srand(FOR);
  int n, k;
  cin >> n >> k;
  if (n - 1 < k) {
    cout << -1;
    return 0;
  }
  int now = n;
  while (k--) {
    a[now] = now;
    now--;
  }
  for (int i = now; i >= 1; i--) {
    a[i] = i - 1;
  }
  a[1] = now;
  for (int i = 1; i <= n; i++) {
    cout << a[i] << " ";
  }
  return 0;
}
