Friends and Cookies(思维)

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。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define LL long long int
 using namespace std;
 int main()
 {
     int t;
     LL i,j,a[],m,n,x,y;
     scanf("%d",&t);
     while(t--)
     {
         memset(a,,sizeof(a));
         scanf("%lld%lld",&m,&n);
         if(n==)///特判一下当n为1的时候
         {
             printf("%lld\n",m);
         }
         else if(m<n)
         {
             for(i=;i<m;i++)
             {
                 a[i]=;
             }
             for(i=;i<n;i++)///只有前m个人有蛋糕，后面的人没有了蛋糕
             {
                 if(i==n-)
                 {
                     printf("%lld\n",a[i]);
                 }
                 else
                 {
                     printf("%lld ",a[i]);
                 }
             }

         }
         else
         {
             for(i=;i<n;i++)///第一行的分配
             {
                 a[i]=;
             }
             m=m-n;///分配完第一行之后剩下的蛋糕数量
             x=m/(n-);///x表示剩下的行数
             y=m%(n-);///y表示最后一行余下的蛋糕份数
             for(i=;i<n-;i++)
             {
                 a[i]=a[i]+x;///除了第一个人和最后一个人其他的人再得到x份蛋糕
             }
             ///下面是对第一个人和最后一个人以及最后一行的蛋糕分配
             if(x%==)///行数为偶数
             {
                 a[]=a[]+x/;
                 a[n-]=a[n-]+x/;
                 for(i=n-;i>=n--y+;i--)
                 {
                     a[i]++;
                 }
             }
             else///行数为奇数
             {
                 a[]=a[]+x/+;
                 a[n-]=a[n-]+x/;
                 for(i=;i<=y;i++)
                 {
                     a[i]=a[i]+;
                 }
             }
         }
         for(i=;i<n;i++)///输出
         {
             if(i==n-)
             {
                 printf("%lld\n",a[i]);
             }
             else
             {
                 printf("%lld ",a[i]);
             }
         }
     }
     return ;
 }