Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int dr[] = {1, 0, -1, 0, -1, -1, 1, 1};
int dc[] = {0, -1, 0, 1, -1, 1, -1, 1};
int arr[1000005];
int dp[1000005];
int dis[100000];
int dp2[1000005];
int ex[1000005];
int vis[1000000];
int n;
long long int num;
map<int, int> mp;
int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int main() {
  int m;
  long long int q = 1, w, e, r, t, y;
  num = 1;
  cin >> n >> m;
  int ok = n;
  if (n == m)
    cout << -1;
  else {
    arr[1] = n;
    for (int i = 2; i <= n; i++) {
      arr[i] = i - 1;
    }
    for (int i = 1; i <= m / 2 && i * 2 + 1 <= n - 1; i++) {
      swap(arr[i * 2 + 1], arr[i * 2 + 2]);
    }
    if (m % 2 && n % 2)
      swap(arr[1], arr[n]);
    else if (m % 2)
      swap(arr[1], arr[2]);
    if (n % 2 && m + 1 == n && n > 1)
      swap(arr[1], arr[n]), swap(arr[1], arr[2]);
    for (int i = 1; i <= n; i++) {
      cout << arr[i] << " ";
    }
  }
}