Drivers Dissatisfaction
Contest: Codeforces Round 378 (Div. 2)
Difficulty rating: 2200
Tags: data structures, dsu, graphs, trees

In one kingdom there are n cities and m two-way roads. Each road connects a pair of cities, and for each road we know the level of drivers dissatisfaction — the value wi.

For each road we know the value ci — how many lamziks we should spend to reduce the level of dissatisfaction with this road by one. Thus, to reduce the dissatisfaction with the i-th road by k, we should spend k·ci lamziks. And it is allowed for the dissatisfaction to become zero or even negative.

In accordance with the king's order, we need to choose n - 1 roads and make them the main roads. An important condition must hold: it should be possible to travel from any city to any other by the main roads.

The road ministry has a budget of S lamziks for the reform. The ministry is going to spend this budget for repair of some roads (to reduce the dissatisfaction with them), and then to choose the n - 1 main roads.

Help to spend the budget in such a way and then to choose the main roads so that the total dissatisfaction with the main roads will be as small as possible. The dissatisfaction with some roads can become negative. It is not necessary to spend whole budget S.

It is guaranteed that it is possible to travel from any city to any other using existing roads. Each road in the kingdom is a two-way road.

Input format:
The first line contains two integers n and m (2 ≤ n ≤ 2·105, n - 1 ≤ m ≤ 2·105) — the number of cities and the number of roads in the kingdom, respectively.

The second line contains m integers w1, w2, ..., wm (1 ≤ wi ≤ 109), where wi is the drivers dissatisfaction with the i-th road.

The third line contains m integers c1, c2, ..., cm (1 ≤ ci ≤ 109), where ci is the cost (in lamziks) of reducing the dissatisfaction with the i-th road by one.

The next m lines contain the description of the roads. The i-th of this lines contain a pair of integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) which mean that the i-th road connects cities ai and bi. All roads are two-way oriented so it is possible to move by the i-th road from ai to bi, and vice versa. It is allowed that a pair of cities is connected by more than one road.

The last line contains one integer S (0 ≤ S ≤ 109) — the number of lamziks which we can spend for reforms.

Output format:
In the first line print K — the minimum possible total dissatisfaction with main roads.

In each of the next n - 1 lines print two integers x, vx, which mean that the road x is among main roads and the road x, after the reform, has the level of dissatisfaction vx.

Consider that roads are numbered from 1 to m in the order as they are given in the input data. The edges can be printed in arbitrary order. If there are several answers, print any of them.

Constraints:
Time limit: 4.0 s
Memory limit: 256.0 MB

Example input:
6 9
1 3 1 1 3 1 2 2 2
4 1 4 2 2 5 3 1 6
1 2
1 3
2 3
2 4
2 5
3 5
3 6
4 5
5 6
7

Example output:
0
1 1
3 1
6 1
7 2
8 -5

Example input:
3 3
9 5 1
7 7 2
2 1
3 1
3 2
2

Example output:
5
3 0
2 5

Editorial:
If you choose any n - 1 roads then price of reducing overall dissatisfaction is equal to min(c1, c2, ..cn - 1) where сi is price of reducing by 1 dissatisfaction of i-th edge. So the best solution is to choose one edge and reduce dissatisfaction of it until running out of budget.
Let's construct minimal spanning tree using Prim or Kruskal algorithm using edges of weights equal to dissatisfaction and calculate minimal price of reducing dissatisfaction. Time complexity — $${\mathcal {O}}(m\log n)$$.
Now we can iterate over edges implying that current is the one to be reduced to minimum. For example, for every edge we can build new MST and recalculate answer. It's $${\mathcal {O}}(m^{2}\cdot \log n)$$. Therefore we should use this fact: it's poinless to reduce dissatisfaction of edges which weren't selected to be main.
Then we can transform original MST instead of constructing m new ones. Add next edge to MST, now it contains a cycle from which edge with maximal dissatisfaction is about to be deleted. This can be achieved in such a way: find LCA of vertices of new edge in $${\mathcal {O}}(\log n)$$ and using binary lifting with precalc in $${\mathcal {O}}(\log n)$$ find the edge to delete.
Time complexity — $$O(n \log n + m \log n)$$.