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>
#pragma warning(disable : 4786)
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const double EPS = 1e-11;
const int INF = 2e30;
const double PI = 2 * acos(0.0);
int MAX(int a, int b) { return a > b ? a : b; }
long long MIN(long long a, long long b) { return a < b ? a : b; }
void SWAP(int &a, int &b) {
  int t = a;
  a = b;
  b = t;
}
int GCD(int a, int b) {
  while (b) {
    b ^= a ^= b ^= a %= b;
  }
  return a;
}
bool isPrime[100005];
int primeCount[100005];
void seive() {
  int i, j;
  for (i = 0; i < 100005; i++) {
    isPrime[i] = true;
    primeCount[i] = 0;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (i = 2; i < 100005; i++) {
    if (isPrime[i] == true) {
      for (j = 2 * i; j < 100005; j += i) {
        isPrime[j] = false;
      }
    }
  }
  for (i = 0, j = 0; i < 100005; i++) {
    if (isPrime[i] == true) j++;
    primeCount[i] = j;
  }
}
int main() {
  seive();
  int i, j, n, k, arr[100005];
  scanf("%d %d", &n, &k);
  if (n == k) {
    printf("-1\n");
  } else {
    for (i = 2; i <= n - k; i++) {
      if (i != 2) printf(" ");
      printf("%d", i);
    }
    printf(" 1");
    for (; i <= n; i++) printf(" %d", i);
    printf("\n");
  }
  return 0;
}
