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;
void sf(int &x) { scanf("%d", &x); }
void sf(long long &x) { scanf("%lld", &x); }
void sf(long long &x, long long &y) { scanf("%lld%lld", &x, &y); }
void sf(float &x) { scanf("%f", &x); }
void sf(double &x) { scanf("%lf", &x); }
void sf(int &x, int &y) { scanf("%d%d", &x, &y); }
void sf(float &x, float &y) { scanf("%f%f", &x, &y); }
void sf(double &x, double &y) { scanf("%lf%lf", &x, &y); }
void sf(double &x, double &y, double &z) { scanf("%lf%lf%lf", &x, &y, &z); }
void sf(int &x, int &y, int &z) { scanf("%d%d%d", &x, &y, &z); }
void sf(long long &x, long long &y, long long &z) {
  scanf("%lld%lld%lld", &x, &y, &z);
}
void sf(float &x, float &y, float &z) { scanf("%u%u%u", &x, &y, &z); }
void sf(char &x) { x = getchar(); }
void sf(char *s) { scanf("%s", s); }
void sf(string &s) { cin >> s; }
void pf(int x) { printf("%d\n", x); }
void pf(int x, int y) { printf("%d %d\n", x, y); }
void pf(int x, int y, int z) { printf("%d %d %d\n", x, y, z); }
void pf(long long x) { printf("%lld\n", x); }
void pf(long long x, long long y) { printf("%lld %lld\n", x, y); }
void pf(long long x, long long y, long long z) {
  printf("%lld %lld %lld\n", x, y, z);
}
void pf(float x) { printf("%u\n", x); }
void pf(double x) { printf("%.6lf\n", x); }
void pf(double x, double y) { printf("%.5lf %.5lf\n", x, y); }
void pf(char x) { printf("%c\n", x); }
void pf(char *x) { printf("%s\n", x); }
void pf(string x) {
  cout << x;
  puts("");
}
long long STN(string s) {
  long long sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
template <class T>
T bigmod(T b, T p, T m) {
  if (p == 0) return 1 % m;
  T x = b;
  T ans = 1;
  while (p) {
    if (p & 1) ans = (ans * x) % m;
    p >>= 1;
    x = (x * x) % m;
  }
  return ans;
}
template <class T>
T gcd(T x, T y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
template <typename T>
T POW(T b, T p) {
  if (p == 0) return 1;
  if (p == 1) return b;
  if (p % 2 == 0) {
    T s = POW(b, p / 2);
    return s * s;
  }
  return b * POW(b, p - 1);
}
template <typename T>
T modinv(T num, T m) {
  return bigmod(num, m - 2, m);
}
template <class T>
string NTS(T Number) {
  stringstream ss;
  ss << Number;
  return ss.str();
}
int n, k, a[100009];
int main() {
  sf(n, k);
  if (n == k)
    pf(-1);
  else {
    if (n & 1) {
      for (int i = 2; i <= n; i += 2) a[i] = i + 1, a[i + 1] = i;
      a[1] = 1;
    } else {
      for (int i = 2; i < n; i += 2) a[i] = i + 1, a[i + 1] = i;
      a[1] = n;
      a[n] = 1;
    }
    int i = 2;
    for (int j = 0; j < k / 2; j++) {
      swap(a[i], a[i + 1]);
      i += 2;
    }
    if (k & 1) {
      if (n & 1) {
        a[n - 1] = n - 1;
        a[1] = n;
        a[n] = 1;
      } else {
        swap(a[1], a[n]);
      }
    }
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
  }
  return 0;
}