Four Vertices
Contest: Codeforces Global Round 16
Difficulty rating: 3100
Tags: constructive algorithms, data structures, graphs, greedy, implementation, shortest paths

You are given an undirected weighted graph, consisting of $$$n$$$ vertices and $$$m$$$ edges.

Some queries happen with this graph:

- Delete an existing edge from the graph.
- Add a non-existing edge to the graph.

At the beginning and after each query, you should find four different vertices $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ such that there exists a path between $$$a$$$ and $$$b$$$, there exists a path between $$$c$$$ and $$$d$$$, and the sum of lengths of two shortest paths from $$$a$$$ to $$$b$$$ and from $$$c$$$ to $$$d$$$ is minimal. The answer to the query is the sum of the lengths of these two shortest paths. The length of the path is equal to the sum of weights of edges in this path.

Input format:
The first line contains two integers $$$n$$$ and $$$m$$$ $$$(4 \le n, m \le 10^5)$$$ — the number of vertices and edges in the graph respectively.

Each of the next $$$m$$$ lines contain three integers $$$v$$$, $$$u$$$, $$$w$$$ ($$$1 \le v, u \le n, v \neq u$$$, $$$1 \le w \le 10^9$$$) — this triple means that there is an edge between vertices $$$v$$$ and $$$u$$$ with weight $$$w$$$.

The next line contains a single integer $$$q$$$ $$$(0 \le q \le 10^5)$$$ — the number of queries.

The next $$$q$$$ lines contain the queries of two types:

- $$$0$$$ $$$v$$$ $$$u$$$ — this query means deleting an edge between $$$v$$$ and $$$u$$$ $$$(1 \le v, u \le n, v \neq u)$$$. It is guaranteed that such edge exists in the graph.
- $$$1$$$ $$$v$$$ $$$u$$$ $$$w$$$ — this query means adding an edge between vertices $$$v$$$ and $$$u$$$ with weight $$$w$$$ ($$$1 \le v, u \le n, v \neq u$$$, $$$1 \le w \le 10^9$$$). It is guaranteed that there was no such edge in the graph.

It is guaranteed that the initial graph does not contain multiple edges.

At the beginning and after each query, the graph doesn't need to be connected.

It is guaranteed that at each moment the number of edges will be at least $$$4$$$. It can be proven, that at each moment there exist some four vertices $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ such that there exists a path between vertices $$$a$$$ and $$$b$$$, and there exists a path between vertices $$$c$$$ and $$$d$$$.

Output format:
Print $$$q + 1$$$ integers — the minimal sum of lengths of shortest paths between chosen pairs of vertices before the queries and after each of them.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
Before the queries you can choose vertices $$$(a, b) = (3, 2)$$$ and $$$(c, d) = (1, 4)$$$. The sum of lengths of two shortest paths is $$$3 + 1 = 4$$$.

After the first query you can choose vertices $$$(a, b) = (2, 5)$$$ and $$$(c, d) = (1, 4)$$$. The sum of lengths of two shortest paths is $$$2 + 1 = 3$$$.

After the second query you can choose vertices $$$(a, b) = (3, 4)$$$ and $$$(c, d) = (2, 5)$$$. The sum of lengths of two shortest paths is $$$1 + 2 = 3$$$.

After the third query, you can choose vertices $$$(a, b) = (2, 6)$$$ and $$$(c, d) = (4, 5)$$$. The sum of lengths of two shortest paths is $$$4 + 3 = 7$$$.

After the last query you can choose vertices $$$(a, b) = (1, 6)$$$ and $$$(c, d) = (2, 5)$$$. The sum of lengths of two shortest paths is $$$3 + 2 = 5$$$.

Example input:
6 6
1 3 6
4 3 1
1 4 1
2 6 4
2 4 2
5 4 3
4
1 2 5 2
0 1 4
0 3 4
1 6 1 3

Example output:
4
3
3
7
5

Editorial:
Observation: the answer always consists either of three edges that have one common end or of two edges that do not have any common ends.
To find the answer of the first type it is enough to maintain three minimal edges for each vertex.
To find the answer of the second type we will use this observation: we can leave in the graph only those edges that are in the set of three minimal edges for each their end. Let's prove that. Let's assume that the answer consists of two edges $$$(a, b)$$$ and $$$(c, d)$$$ and there are at least three edges $$$(a, a_1)$$$, $$$(a, a_2)$$$, $$$(a, a_3)$$$ less than $$$(a, b)$$$. Then the edge $$$(a, b)$$$ can be swapped by one of these edges because at least one of te integers $$$a_1$$$, $$$a_2$$$, $$$a_3$$$ is not equal to $$$c$$$ and $$$d$$$. Then we will maintain the set of all these edges.
Now let's consider two cases. Let $$$(a, b)$$$ be the minimal edge. If it is in the answer then we need to find the minimal edge, that does not have any common vertices with $$$(a, b)$$$. In this case there are at most $$$6$$$ "bad" edges because the degrees of each vertex in the graph that consists only of remaining edges do not exceed $$$3$$$. It means that we have to consider $$$O(1)$$$ variants. If $$$(a, b)$$$ is not in the answer, then there are two edges in the answer that have vertices $$$a$$$ and $$$b$$$ as one of their ends. But there are at most $$$9$$$ such pairs of edges, so we have to consider only $$$O(1)$$$ variants.
So the final answer is the minimum of answers of both types.