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;
vector<int> niz;
int ostatak = 0;
int n, k;
bool netacan = false;
int gcd(int a, int b) {
  int c;
  while (a != 0) {
    c = a;
    a = b % a;
    b = c;
  }
  return b;
}
bool funkc(int i, int j) {
  int prije = 0, posle = 0;
  if (gcd(i, niz[i]) > 1) {
    prije++;
  }
  if (gcd(j, niz[j]) > 1) {
    prije++;
  }
  if (gcd(j, niz[i]) > 1) {
    posle++;
  }
  if (gcd(i, niz[j]) > 1) {
    posle++;
  }
  int rez = posle - prije;
  if (rez > 0 && rez <= ostatak) {
    ostatak -= rez;
    int t = niz[i];
    niz[i] = niz[j];
    niz[j] = t;
    return true;
  }
  return false;
}
void skidaj() {
  bool promjena = true;
  while (promjena) {
    promjena = false;
    for (int i = 0; i < n; i++) {
      if (ostatak == 0) {
        break;
      }
      for (int j = 0; j < n; j++) {
        if (ostatak == 0) {
          break;
        }
        if (i == j) {
          continue;
        }
        bool f = funkc(i, j);
        if (f) {
          promjena = true;
        }
      }
    }
  }
  if (ostatak > 0) {
    netacan = true;
  }
}
int main() {
  scanf("%d", &n);
  scanf("%d", &k);
  if (n == k) {
    netacan = true;
  } else {
    for (int i = 1; i <= n; i++) {
      niz.push_back(i);
    }
    ostatak = n - k - 1;
    if (ostatak > 0) {
      skidaj();
    }
  }
  if (netacan) {
    printf("-1");
  } else {
    for (int i = 0; i < n; i++) {
      printf("%d ", niz[i]);
    }
  }
}
