DreamGrid, the king of Gridland, is making a knight tournament. There are knights, numbered from 1 to , participating in the tournament. The rules of the tournament are listed as follows:

• The tournament consists of rounds. Each round consists of several duels. Each duel happens between exactly two knights.
• Each knight must participate in exactly one duel during each round.
• For each pair of knights, there can be at most one duel between them during all the rounds.
• Let , , and , be four distinct integers. If
  • Knight fights against knight during round , and
  • Knight fights against knight during round , and
  • Knight fights against knight during round ,
  then knight must fight against knight during round .

As DreamGrid's general, you are asked to write a program to arrange all the duels in all the rounds, so that the resulting arrangement satisfies the rules above.

Input

There are multiple test cases. The first line of the input is an integer , indicating the number of test cases. For each test case:

The first and only line contains two integers and (), indicating the number of knights participating in the tournament and the number of rounds.

It's guaranteed that neither the sum of nor the sum of in all test cases will exceed 5000.

Output

For each test case:

• If it's possible to make a valid arrangement, output lines. On the -th line, output integers separated by one space, indicating that in the -th round, knight will fight against knight for all .

  If there are multiple valid answers, output the lexicographically smallest answer.

  Consider two answers and , let's denote as the -th integer on the -th line in answer , and as the -th integer on the -th line in answer . Answer is lexicographically smaller than answer , if there exists two integers () and (), such that

  • for all and , , and
  • for all , , and finally .

   

• If it's impossible to make a valid arrangement, output "Impossible" (without quotes) in one line.

Please, DO NOT output extra spaces at the end of each line, or your answer may be considered incorrect!

Sample Input

2
3 1
4 3

Sample Output

Impossible
2 1 4 3
3 4 1 2
4 3 2 1

題解：我們另第一行為1 2 3 4 5 6 7 8 ... 寫出幾個可以看出

1 2 3 4 5 6 7 8

2 1 4 3 6 5 8 7

3 4 1 2 7 8 5 6

4 3 2 1 8 7 6 5

5 6 7 8 1 2 3 4

6 5 8 7 2 1 3 4  寫到這裡是不是就可以看出規律了 就是 先是兩個互換 然後接下來 兩個 對應上面左半部分和右半部分的逆序

最後4個為 上4個的逆序  然後對應一個n  我們可以進行多少次戰爭呢 就是 lowbit(n)  - 1

#include <iostream>
#include<cstdio>
using namespace std;
int n,k;
int dp[1100][1100];
void dfs(int l,int r,int cur)
{
    if(cur==1)
    {
        for(int i=1;i<=n;i++)
            dp[cur][i]=i;
        return;
    }
    dfs(l,(l+r)/2,cur/2);
    for(int i=cur/2+1;i<=cur;i++)
    {
        for(int j=1;j<=n;j+=cur)
        {
            int k=j+cur-1;
            for(int mm=j;mm<j+cur;mm++)
            {
                dp[i][mm]=dp[cur/2-(i-cur/2-1)][k-(mm-j)];
            }

        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&k);
        int m=n;
        int flag=1;
        int cnt=0;
        while((m&1) != 1)
        {
            cnt++;
            m>>=1;
        }
        if(flag&&(1<<cnt)-1<k) flag=0;
        if(!flag) printf("Impossible\n");
        else
        {
            dfs(1,n,(1<<cnt));
            for(int i=2;i<=k+1;i++)
            {
                for(int j=1;j<=n;j++)
                    printf("%d%c",dp[i][j]," \n"[j==n]);
            }
        }
    }
    return 0;
}