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## Magical Candy Store november long challenge problem solution code with explaination-

Magical Candy Store November long challenge problem solution code with explanation

Chef and Chefu are at a magical candy store playing a game with the following rules:

• There are two candy counters; each of them stores an infinite number of candies. At any time, only one of the counters is open and the other is closed.
• Exactly one player is present at each of the counters. Initially, Chef is at the open counter and Chefu is at the closed counter.
• There is a sequence of $N$ distinct integers ${A}_{1},{A}_{2},\dots ,{A}_{N}$. The game consists of $R$ turns; in the $i$-th turn, the open counter offers only $C={A}_{\left(i-1\right)\mathrm{%}N+1}$ candies to the player present at this counter. This player should choose a positive number of candies $M$ to accept, where $1\le M\le C$.
• If this player accepts an odd number of candies, the players have to swap their positions (each player goes to the other counter).
• After each $N$ turns, the counter which was currently open is closed and the counter which was currently closed is opened.
• The primary goal of each player is to maximise his own number of candies after $R$ turns. As a second priority, each player wants to minimise the number of candies his opponent has after $R$ turns.

You should process $Q$ queries. In each query, you are given $R$ and you should find the number of candies Chef has after $R$ turns, assuming that both players play the game optimally. Since this number could be very large, compute it modulo ${10}^{9}+7$.

### 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 line of each test case contains a single integer $N$.
• The second line contains $N$ space-separated integers ${A}_{1},{A}_{2},\dots ,{A}_{N}$.
• The third line contains a single integer $Q$.
• Each of the next $Q$ lines contains a single integer $R$ describing a query.

### Output

For each query, print a single line containing one integer ― the maximum number of candies Chef can get, modulo ${10}^{9}+7$.

### Constraints

• $1\le T\le 25$
• $1\le N\le {10}^{5}$
• $1\le {A}_{i}\le {10}^{9}$ for each valid $i$
• ${A}_{1},{A}_{2},\dots ,{A}_{N}$ are pairwise distinct
• $1\le Q\le {10}^{5}$
• $1\le R\le {10}^{12}$
• the sum of $N+Q$ over all test cases does not exceed ${10}^{6}$

• $N\le 10$
• $Q\le 35$
• $R\le 35$

Subtask #2 (85 points): original constraints

### Example Input

1
4
4 10 2 1
2
4
5


### Example Output

17
21


### Explanation

Example case 1: In the $1$-st, $2$-nd and $3$-rd turn, Chef takes $4$$10$ and $2$ candies ($16$ in total) respectively. In the $4$-th turn, Chef takes $1$ candy ($17$ in total; this is the answer to the first query), which is odd and hence he has to go to the counter which is closed. However, since $N=4$ turns are just completed, the counter which was currently open closes and the other one (where Chef went) opens. In the $5$-th round, Chef can take $4$ candies, so he has $21$ candies.

### Solution-

Must Join Telegram channel for Editorial -

Hint-
Here is major role of element 1.

for input-
3
5
1 2 3 4 5
7
1
2
3
5
6
10
11
5
2 3 4 5 1
10
1
2
3
5
6
7
10
11
20
21
5
2 3 1 4 5
12
1
2
3
4
5
6
7
8
9
10
11
12
output should be-
1
1
1
1
1
2
2
2
5
8
13
15
18
26
28
52
54
2
5
5
9
14
16
19
19
23
28
30
33

### Solution -

I hope you read the hint actually this question is totally based on logical thinking.
once you get the logic it is very easy to code.
Here are many cases which you have to think and handle them.
First of all, assume that there is no 1 in the given sequence since the chef is so cleaver  this means that the chef  never want to handover any turn to chefu.

The chef always try to handover his turn to chefu if anytime 1 occurred. because for element 1 there is no selection choice left for chef and chefu.
Since Chef is playing first so he will choose even number and he will choose odd when next element is 1. so that chefu have to choose 1 because he has no choice left.

We can make 4 cases here and ans can differ according to the poison of 1.

Case 1- if No 1 present.
Case 2- if 1 present at the beginning.
case 3- if 1 present at the mid.
case 4 if 1 present at the last.

If you are able to handle these cases successfully you can solve this question easily.

To perform this task more effectively make a prefix array.And chose always even except the last position.

For Video editorial follow video below-

Now see my code-