Codeforce Round #210 Div2 B

B. Levko and Permutation

time limit per test 1 second

memory limit per test 256 megabytes

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 p_i of permutation p₁, p₂, ... , p_n is good if gcd(i, p_i) > 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.

Sample test(s)

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.

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 10055
#define ll long long
#define mod 1000000007
#define INF 0x7fffffff
#define eps 1e-8
int n, m, k, sx, sy;
int a[maxn];
int main(){
    int t;
    //scanf("%d", &t);
    //while (t--){
    //memset(a, 0, sizeof a);
    //memset(b, 0, sizeof b);
    scanf("%d%d", &n, &m);
    if (m == n){ printf("-1\n"); return 0; }
    for (int i = 1; i <= n; i++){
            if (i < n - m)printf("%d ", i + 1);
            if (i == n - m)printf("1 ");
            if (i>n - m)printf("%d ", i);
        }
        printf("\n");
    //}
    return 0;
}