Friends and Cookies

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<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 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)
20         {
21             for(i=0;i<m;i++)
22             {
23                 a[i]=1;
24             }
25             for(i=0;i<n;i++)///只有前m个人有蛋糕，后面的人没有了蛋糕
26             {
27                 if(i==n-1)
28                 {
29                     printf("%lld\n",a[i]);
30                 }
31                 else
32                 {
33                     printf("%lld ",a[i]);
34                 }
35             }
36 
37         }
38         else
39         {
40             for(i=0;i<n;i++)///第一行的分配
41             {
42                 a[i]=1;
43             }
44             m=m-n;///分配完第一行之后剩下的蛋糕数量
45             x=m/(n-1);///x表示剩下的行数
46             y=m%(n-1);///y表示最后一行余下的蛋糕份数
47             for(i=1;i<n-1;i++)
48             {
49                 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
50             }
51             ///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
52             if(x%2==0)///行数为偶数
53             {
54                 a[0]=a[0]+x/2;
55                 a[n-1]=a[n-1]+x/2;
56                 for(i=n-2;i>=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<n;i++)///输出
72         {
73             if(i==n-1)
74             {
75                 printf("%lld\n",a[i]);
76             }
77             else
78             {
79                 printf("%lld ",a[i]);
80             }
81         }
82     }
83     return 0;
84 }