Weights Division (hard version)
Contest: Codeforces Round 661 (Div. 3)
Difficulty rating: 2200
Tags: binary search, dfs and similar, greedy, sortings, trees, two pointers

Easy and hard versions are actually different problems, so we advise you to read both statements carefully.

You are given a weighted rooted tree, vertex $$$1$$$ is the root of this tree. Also, each edge has its own cost.

A tree is a connected graph without cycles. A rooted tree has a special vertex called the root. A parent of a vertex $$$v$$$ is the last different from $$$v$$$ vertex on the path from the root to the vertex $$$v$$$. Children of vertex $$$v$$$ are all vertices for which $$$v$$$ is the parent. A vertex is a leaf if it has no children. The weighted tree is such a tree that each edge of this tree has some weight.

The weight of the path is the sum of edges weights on this path. The weight of the path from the vertex to itself is $$$0$$$.

You can make a sequence of zero or more moves. On each move, you select an edge and divide its weight by $$$2$$$ rounding down. More formally, during one move, you choose some edge $$$i$$$ and divide its weight by $$$2$$$ rounding down ($$$w_i := \left\lfloor\frac{w_i}{2}\right\rfloor$$$).

Each edge $$$i$$$ has an associated cost $$$c_i$$$ which is either $$$1$$$ or $$$2$$$ coins. Each move with edge $$$i$$$ costs $$$c_i$$$ coins.

Your task is to find the minimum total cost to make the sum of weights of paths from the root to each leaf at most $$$S$$$. In other words, if $$$w(i, j)$$$ is the weight of the path from the vertex $$$i$$$ to the vertex $$$j$$$, then you have to make $$$\sum\limits_{v \in leaves} w(root, v) \le S$$$, where $$$leaves$$$ is the list of all leaves.

You have to answer $$$t$$$ independent test cases.

Input format:
The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of the test case contains two integers $$$n$$$ and $$$S$$$ ($$$2 \le n \le 10^5; 1 \le S \le 10^{16}$$$) — the number of vertices in the tree and the maximum possible sum of weights you have to obtain. The next $$$n-1$$$ lines describe edges of the tree. The edge $$$i$$$ is described as four integers $$$v_i$$$, $$$u_i$$$, $$$w_i$$$ and $$$c_i$$$ ($$$1 \le v_i, u_i \le n; 1 \le w_i \le 10^6; 1 \le c_i \le 2$$$), where $$$v_i$$$ and $$$u_i$$$ are vertices the edge $$$i$$$ connects, $$$w_i$$$ is the weight of this edge and $$$c_i$$$ is the cost of this edge.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$10^5$$$ ($$$\sum n \le 10^5$$$).

Output format:
For each test case, print the answer: the minimum total cost required to make the sum of weights paths from the root to each leaf at most $$$S$$$.

Constraints:
Time limit: 3.0 s
Memory limit: 256.0 MB

Example input:
4
4 18
2 1 9 2
3 2 4 1
4 1 1 2
3 20
2 1 8 1
3 1 7 2
5 50
1 3 100 1
1 5 10 2
2 3 123 2
5 4 55 1
2 100
1 2 409 2

Example output:
0
0
11
6

Editorial:
Read the easy version editorial first, because almost all solution is the solution to the easy version with little changes.
Firstly, let's simulate the greedy process we used to solve the easy version problem, for edges with cost $$$1$$$ and edges with cost $$$2$$$, independently. But we don't stop when our sum reach something, let's simulate until our sum becomes $$$0$$$ and store each intermediate result in the array $$$v_1$$$ for edges with cost $$$1$$$ and $$$v_2$$$ for edges with cost $$$2$$$. So, the array $$$v_1$$$ contains the initial total impact of edges of cost $$$1$$$, then the impact after making one move, two moves, and so on... The same with $$$v_2$$$ but for edges with cost $$$2$$$.
Now let's fix how many moves on edges with cost $$$1$$$ we do. Let it be $$$i$$$ (and arrays $$$v_1$$$ and $$$v_2$$$ are $$$0$$$-indexed). Then the sum we obtain from the $$$1$$$-cost edges is $$$v_1[i]$$$. So we need to find the minimum number of moves $$$p$$$ we can do on $$$2$$$-cost edges so that $$$v_1[i] + v_2[p] \le S$$$. This can be done using binary search or moving pointer (if we iterate over $$$i$$$ in increasing order, place $$$p$$$ at the end of $$$v_2$$$ and move it to the left while $$$v_1[i] + v_2[p] \le S$$$). Then, if $$$v_1[i] + v_2[p] \le S$$$, we can update the answer with the value $$$i + 2p$$$.
Time complexity is actually the same as in easy version of the problem: $$$O(n \log{w} \log{n})$$$.