Teleporters (Hard Version)
Contest: Codeforces Round 849 (Div. 4)
Difficulty rating: 1900
Tags: binary search, greedy, sortings

The only difference between the easy and hard versions are the locations you can teleport to.

Consider the points $$$0,1,\dots,n+1$$$ on the number line. There is a teleporter located on each of the points $$$1,2,\dots,n$$$. At point $$$i$$$, you can do the following:

- Move left one unit: it costs $$$1$$$ coin.
- Move right one unit: it costs $$$1$$$ coin.
- Use a teleporter at point $$$i$$$, if it exists: it costs $$$a_i$$$ coins. As a result, you can choose whether to teleport to point $$$0$$$ or point $$$n+1$$$. Once you use a teleporter, you can't use it again.

You have $$$c$$$ coins, and you start at point $$$0$$$. What's the most number of teleporters you can use?

Input format:
The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case contains two integers $$$n$$$ and $$$c$$$ ($$$1 \leq n \leq 2\cdot10^5$$$; $$$1 \leq c \leq 10^9$$$)  — the length of the array and the number of coins you have respectively.

The following line contains $$$n$$$ space-separated positive integers $$$a_1,a_2,\dots,a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the costs to use the teleporters.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output format:
For each test case, output the maximum number of teleporters you can use.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first test case, you can move one unit to the right, use the teleporter at index $$$1$$$ and teleport to point $$$n+1$$$, move one unit to the left and use the teleporter at index $$$5$$$. You are left with $$$6-1-1-1-1 = 2$$$ coins, and wherever you teleport, you won't have enough coins to use another teleporter. You have used two teleporters, so the answer is two.

In the second test case, you go four units to the right and use the teleporter to go to $$$n+1$$$, then go three units left and use the teleporter at index $$$6$$$ to go to $$$n+1$$$, and finally, you go left four times and use the teleporter. The total cost will be $$$4+6+3+4+4+9 = 30$$$, and you used three teleporters.

In the third test case, you don't have enough coins to use any teleporter, so the answer is zero.

Example input:
10
5 6
1 1 1 1 1
8 32
100 52 13 6 9 4 100 35
1 1
5
4 5
4 3 2 1
5 9
2 3 1 4 1
5 8
2 3 1 4 1
4 3
2 3 4 1
4 9
5 4 3 3
2 14
7 5
5 600000000
500000000 400000000 300000000 200000000 100000000

Example output:
2
3
0
1
3
2
1
1
2
2

Editorial:
Please also refer to the tutorial for the easy version.
If we are not at the first taken portal, the problem is still independent for each portal, but this time the cost of a portal is $$$min(a_i + i, a_i + n + 1 - i)$$$ (since we can come to a portal either from point $$$0$$$ or point $$$n+1$$$). So, we again sort the portals by their costs. But this time, we need to make sure that the first taken portal is taken from point $$$0$$$, so we will iterate over all portals and check the maximum amount of portals we can take if we use it as the first one. We can check this using prefix sums over the minimum cost array and binary searching, checking if the amount of considered portals taken doesn't exceed the number of coins we initially have (we also have to deal with the case when the portal we are considering is included both times as the initial portal and in the minimum cost prefix).