Abood's birthday has come, and his n friends are aligned in a single line from 1 to n, waiting for their cookies, Abood has x cookies to give to his friends.

Here is an example to understand how Abood gives away the cookies. Suppose Abood has 4 friends and x cookies, then Abood will do the following:

1. Give a cookie to the 1^st friend.
2. Give a cookie to the 2^nd friend.
3. Give a cookie to the 3^rd friend.
4. Give a cookie to the 4^th friend.
5. Give a cookie to the 3^rd friend.
6. Give a cookie to the 2^nd friend.
7. Give a cookie to the 1^st friend.
8. Give a cookie to the 2^nd friend.
9. And so on until all the x cookies are given away.

Your task is to find how many cookies each friend will get. Can you?

Input

The first line contains an integer T (1 ≤ T ≤ 100) specifying the number of test cases.

Each test case consists of a single line containing two integers x and n (1 ≤ x ≤ 10¹⁸, 1 ≤ n ≤ 1000), in which x is the number of cookies Abood has, and n is the number of his friends.

Output

For each test case, print a single line containing n space-separated integers a₁, ..., a_n, in which a_i represents how many cookies the i^th friend got.

Example

Input

1 5 3

Output

2 2 1 解题思路：我将蛋糕的分配方式模拟一下 －　－　－　－　－　－　－ －　－　－　－　－　－ 　　－　－　－　－　－　－ －　－　－　－　－　－　 　　－　－　－　－　－　－ 　　　　　　－　－　－

我们可以看到的是：第一行每一个人都有，然后开始的奇数个前N-1个有，偶数个后N-1有，最后一行再判断是第奇数个还是偶数个就可以得出结果。

注意的是：没跑完第一行的情况需要特判一下,跑完第一行但是m<n也需要特判一下。

1 #include
        
          2 #include
         
           3 #include
          
            4 #define LL long long int 5 using namespace std; 6 int main() 7 { 8     int t; 9     LL i,j,a[10010],m,n,x,y;10     scanf("%d",&t);11     while(t--)12     {13         memset(a,0,sizeof(a));14         scanf("%lld%lld",&m,&n);15         if(n==1)///特判一下当n为1的时候16         {17             printf("%lld\n",m);18         }19         else if(m
           
            =n-2-y+1;i--)57                 {58                     a[i]++;59                 }60             }61             else///行数为奇数62             {63                 a[0]=a[0]+x/2+1;64                 a[n-1]=a[n-1]+x/2;65                 for(i=1;i<=y;i++)66                 {67                     a[i]=a[i]+1;68                 }69             }70         }71         for(i=0;i