## Problem Statement -

this Problem Positive prefix taken from CodeChef December long challenge.Let's read Problem statement.

You are given two positive integers $N$ and $K$, where $K\le N$. Find a sequence ${A}_{1},{A}_{2},\dots ,{A}_{N}$ such that:

• for each valid $i$${A}_{i}$ is either $i$ or $-i$
• there are exactly $K$ values of $i$ such that $1\le i\le N$ and ${A}_{1}+{A}_{2}+\dots +{A}_{i}>0$

If there are multiple solutions, you may print any one of them. It can be proved that at least one solution always exists.

### Input

• The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
• The first and only line of each test case contains two space-separated integers $N$ and $K$.

### Output

For each test case, print a single line containing $N$ space-separated integers ${A}_{1},{A}_{2},\dots ,{A}_{N}$.

### Constraints

• $1\le T\le 1,000$
• $1\le K\le N\le 1,000$

Subtask #1 (10 points): $N\le 10$

Subtask #2 (90 points): original constraints

### Example Input

1
3 3


### Example Output

1 2 3

### Solution-

Hint- learn Prefix array to solve this question.
This question is totally observation-based.You should  good observer to solve  this problem.
Solution since there can be multiple solution of this problem one of them is to print alternate +i and -i.
in my case what i am doing
I have created an output array in which i am storing +i and -i alternatively.
then I iterate over the array and I calculate the prefix array.
Then I calculated total positive and negative number store in the output array.And i build the following logic
if(k<=positive){
int diff=positive-k;
for(int i=n;i>0&&diff>0;i--){
if(prefix[i]>0){
output[i]=-output[i];
diff--;
}

}
}
else{
int diff=k-positive;
for(int i=n;i>0&&diff>0;i--){
if(prefix[i]<0){
output[i]=-output[i];
diff--;
}

}
}

Here is my entire code.

Join Telegram Channel for solution.

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