---
category_name: easy
problem_code: CHEFCOUN
problem_name: 'Counter Test For CHEFSUM'
languages_supported:
- ADA
- ASM
- BASH
- BF
- C
- 'C99 strict'
- CAML
- CLOJ
- CLPS
- 'CPP 4.3.2'
- 'CPP 6.3'
- CPP14
- CS2
- D
- ERL
- FORT
- FS
- GO
- HASK
- ICK
- ICON
- JAVA
- JS
- kotlin
- 'LISP clisp'
- 'LISP sbcl'
- LUA
- NEM
- NICE
- NODEJS
- 'PAS fpc'
- 'PAS gpc'
- PERL
- PERL6
- PHP
- PIKE
- PRLG
- PYPY
- PYTH
- 'PYTH 3.5'
- RUBY
- rust
- SCALA
- 'SCM chicken'
- 'SCM guile'
- 'SCM qobi'
- ST
- swift
- TCL
- TEXT
- WSPC
max_timelimit: '1'
source_sizelimit: '50000'
problem_author: admin2
problem_tester: alex_2oo8
date_added: 1-09-2017
tags:
- admin2
- oct17
- simple
editorial_url: 'https://discuss.codechef.com/problems/CHEFCOUN'
time:
view_start_date: 1508146200
submit_start_date: 1508146200
visible_start_date: 1508146200
end_date: 1735669800
current: 1514815996
layout: problem
---
All submissions for this problem are available.### Read problems statements in [mandarin chinese](http://www.codechef.com/download/translated/OCT17/mandarin/CHEFCOUN.pdf) and [vietnamese](http://www.codechef.com/download/translated/OCT17/vietnamese/CHEFCOUN.pdf) as well.
Once Chef was writing test data for the problem [CHEFSUM](https://codechef.com/problems/CHEFSUM). For your convenience, the summary of this problem is provided as below.
You are given an array **a** of size **n**. Let **prefSum\[i\]** denote the sum of first **i** elements and **sufSum\[i\]** denote the sum of last **n - i + 1** elements of the array **a**. You have to find the least index **i** such that value of **prefSum\[i\] + sufSum\[i\]** is the minimum possible. The bounds/constraints on **n** could be as large as 105.
A newbie programmer was trying to solve this problem. He didn't take care of the fact that the values of **prefSum\[i\] + sufSum\[i\]** might not fit into _unsigned int_ data type. He wrote the following C++ code to solve the problem.
<pre>
<pre>int wrongSolver(std::vector <unsigned int> a) {
int n = a.size();
std::vector<unsigned int> prefSum(n), sufSum(n);
prefSum[0] = a[0];
for (int i = 1; i = 0; i--) {
sufSum[i] = sufSum[i + 1] + a[i];
}
unsigned int mn = prefSum[0] + sufSum[0];
int where = 1;
for (int i = 1; i
</pre>
</pre>
Assume that an _unsigned int_ is 4 bytes long, i.e. it stores values from 0 up to **232 - 1**. Addition of two unsigned int's **x** and **y** is done as **(x + y) modulo 232**. This way, you can see that whenever the value of an unsigned int exceeds the maximum possible value (**232 - 1**), it wraps around.
Chef as a problem setter knows that the above program should not get an AC. Hence, he wants to generate a counter case to fail this solution. He asks your help in generating such a counter case.
### Input
The first line of the input contains an integer **T** denoting the number of test cases.
The only line of each test case contains a single integer **n** denoting the number of integers in the array **a**.
### Output
For each test case, output **n** space separated integers in a line denoting the content of array **a** for which the above program will give a wrong answer.
### Constraints
- **1** ≤ **T** ≤ **10**
### Subtasks
- **Subtask #1 : (50 points)** **99991** ≤ **n** ≤ **105**, **1** ≤ **ai** ≤ **2 \* 109**
- **Subtask #2 : (50 points)** **99991** ≤ **n** ≤ **105**, **1** ≤ **ai** ≤ **105**