Codeforces Round #655 (Div. 2) A. Omkar and Completion

A. Omkar and Completion

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You have been blessed as a child of Omkar. To express your gratitude, please solve this problem for Omkar!

An array aa of length nn is called complete if all elements are positive and don't exceed 10001000, and for all indices xx,yy,zz (1≤x,y,z≤n1≤x,y,z≤n), ax+ay≠azax+ay≠az (not necessarily distinct).

You are given one integer nn. Please find any complete array of length nn. It is guaranteed that under given constraints such array exists.

Input

Each test contains multiple test cases. The first line contains tt (1≤t≤10001≤t≤1000)  — the number of test cases. Description of the test cases follows.

The only line of each test case contains one integer nn (1≤n≤10001≤n≤1000).

It is guaranteed that the sum of nn over all test cases does not exceed 10001000.

Output

For each test case, print a complete array on a single line. All elements have to be integers between 11 and 10001000 and for all indices xx,yy,zz (1≤x,y,z≤n1≤x,y,z≤n) (not necessarily distinct), ax+ay≠azax+ay≠az must hold.

If multiple solutions exist, you may print any.

Example

input

Copy

2
5
4

output

Copy

1 5 3 77 12
384 384 44 44

Note

It can be shown that the outputs above are valid for each test case. For example, 44+44≠38444+44≠384.

Below are some examples of arrays that are NOT complete for the 1st test case:

[1,2,3,4,5][1,2,3,4,5]

Notice that a1+a2=a3a1+a2=a3.

[1,3000,1,300,1][1,3000,1,300,1]

Notice that a2=3000>1000a2=3000>1000.

题意：给你一个n，输出保证每个元素不超过1000的n个元素，并且两元素之和不等于第三元素

解题思路：拿到题目，可能很多人的第一想法就是枚举，其实这是一个实打实的水题，算是本场的签到题吧，我们仔细分析就能得到，每个元素都相等且小于1k，那就满足题意了，人家没说元素不能重复，所以直接打印n个相同的元素就行。

代码如下：

#include<cstdio>
int t,n;int main(void)
{
    
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n-1;++i)
        {
            printf("1 ");
        }
        puts("1");
    }
}