B. Taisia and Dice

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Taisia has n� six-sided dice. Each face of the die is marked with a number from 11 to 66, each number from 11 to 66 is used once.

Taisia rolls all n� dice at the same time and gets a sequence of values a1,a2,…,an�1,�2,…,�� (1≤ai≤61≤��≤6), where ai�� is the value on the upper face of the i�-th dice. The sum of this sequence is equal to s�.

Suddenly, Taisia's pet cat steals exactly one dice with maximum value ai�� and calculates the sum of the values on the remaining n−1�−1 dice, which is equal to r�.

You only know the number of dice n� and the values of s�, r�. Restore a possible sequence a� that fulfills the constraints.

Input

The first line contains the integer t� (1≤t≤10001≤�≤1000) — the number of testcases.

Each testcase is given on a separate line and contains three integers n�, s�, r� (2≤n≤502≤�≤50, 1≤r

It is guaranteed that a solution exists.

Output

For each testcase, print: n� integers a1,a2,…,an�1,�2,…,�� in any order. It is guaranteed that such sequence exists.

If there are multiple solutions, print any.

Example

input

Copy

7

2 2 1

2 4 2

4 9 5

5 17 11

3 15 10

4 4 3

5 20 15

output

Copy

1 1
2 2 
1 2 2 4
6 4 2 3 2
5 5 5
1 1 1 1
1 4 5 5 5

解题说明：此题是一道数学题，有 n个色子，它们朝上的面的点数之和为 s，去掉其中一个色子后的和为 r，问原来所有色子的点数。其中有一个棋子为 s − r，则题目转换为求 n − 1个不大于 6的数且和为 r。

#include
#include
int main()
{
	int t;
	scanf("%d", &t);
	for (int i = 0; i < t; i++)
	{
		int n, s, r;
		scanf("%d%d%d", &n, &s, &r);
		printf("%d ", s - r);
		for (int j = 1; j < n; j++)
		{
			printf("%d ", (r / (n - j)));
			r = r - (r / (n - j));
		}
		printf("\n");
	}
	return 0;
}