B. Levko and Permutation 简单构造

B. Levko and Permutation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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.

const int INF = 1000000000;
const double eps = 1e-8;
const int maxn = 300000;
int ans[maxn];
int main() 
{
    //freopen("in.txt","r",stdin);
    int n,k;
    while(cin>>n>>k)
    {
        if(k > n - 1)
        {
            cout<<-1<<endl;
            continue;
        }
        repf(i,1,n)
            ans[i] = i;
        int Max = n - 1;
        int temp = Max - k;
        
        int t = temp/2;
        int e;
        for(int i = n;i>=2;i-=2)
        {
            if(t)
            {
                swap(ans[i],ans[i-1]);
                t--;
            }else
            {
               e = i; 
            }
        }
        if(temp%2)
        {
            swap(ans[1],ans[e]);
        }
        
        printf("%d",ans[1]);
        repf(i,2,n)
            printf(" %d",ans[i]);
        cout<<endl;
    }
    return 0;
}