Codeforces Round #166 (Div. 2)C. Secret（构造）

C. Secret

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Greatest Secret Ever consists of n words, indexed by positive integers from 1 to n. The secret needs dividing between k Keepers (let's index them by positive integers from 1 to k), the i-th Keeper gets a non-empty set of words with numbers from the set U_i = (u_i, 1, u_i, 2, ..., u_i, |Ui|). Here and below we'll presuppose that the set elements are written in the increasing order.

We'll say that the secret is safe if the following conditions are hold:

• for any two indexes i, j (1 ≤ i < j ≤ k) the intersection of sets U_i and U_j is an empty set;
• the union of sets U₁, U₂, ..., U_k is set (1, 2, ..., n);
• in each set U_i, its elements u_i, 1, u_i, 2, ..., u_i, |Ui| do not form an arithmetic progression (in particular, |U_i| ≥ 3 should hold).

Let us remind you that the elements of set (u₁, u₂, ..., u_s) form an arithmetic progression if there is such number d, that for all i (1 ≤ i < s) fulfills u_i + d = u_i + 1. For example, the elements of sets (5), (1, 10) and (1, 5, 9) form arithmetic progressions and the elements of sets (1, 2, 4) and (3, 6, 8) don't.

Your task is to find any partition of the set of words into subsets U₁, U₂, ..., U_k so that the secret is safe. Otherwise indicate that there's no such partition.

Input

The input consists of a single line which contains two integers n and k (2 ≤ k ≤ n ≤ 10⁶) — the number of words in the secret and the number of the Keepers. The numbers are separated by a single space.

Output

If there is no way to keep the secret safe, print a single integer "-1" (without the quotes). Otherwise, print n integers, the i-th of them representing the number of the Keeper who's got the i-th word of the secret.

If there are multiple solutions, print any of them.

Sample test(s)

Input

11 3

Output

3 1 2 1 1 2 3 2 2 3 1

Input

5 2

Output

-1

如果n<3*k, 无解，因为此时总有至少一个keeper至多拥有2个Words，构成等差数列，不合题意。
否则，可按以下方法分配n个word给k个keeper，a[i]=j表示第i个word分配给第j个keeper。
a[1]=a[2]=a[2k+1]=1;
a[3]=[4]=a[2k+2]=2;
...
a[2k-1]=a[2k]=a[3k]=k;
如果word有余，则令
a[3k+1]=a[3k+2]=...=1;

AC Code:

#include <iostream>
#include <cstdio>

using namespace std;

int a[1000002], n, k;

int main()
{
    while(scanf("%d %d", &n, &k) != EOF)
    {
        if(n < 3*k) puts("-1");
        else
        {
            int i, j;
            for(i = 2, j = 1; j <= k; i += 2, j++)
                a[i-1] = a[i] = a[2 * k + i / 2] = j;
            for(i = 3 * k + 1; i <= n; i++)
                a[i] = 1;
            printf("%d", a[1]);
            for(int i = 2; i <= n; i++) printf(" %d", a[i]);
            puts("");
        }
    }
    return 0;
}