Problem: 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

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

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. 

Examples

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


Output

0
1 1
3 1
6 1
7 2
8 -5


Input

3 3
9 5 1
7 7 2
2 1
3 1
3 2
2


Output

5
3 0
2 5

Solution (in JAVA):

Solution:
import java.util.*;
import java.util.stream.*;
import java.io.*;
import java.math.*;
import java.awt.geom.*;
public class Main {
  public static final BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
  public static final PrintWriter outWriter = new PrintWriter(System.out);
  public static void main() {
    int n = nextInt(), m = nextInt();
    Node[] nodes = new Node[n];
    for (int i = 0; i < nodes.length; i++) { nodes[i] = new Node(i); }
    int[] ws = new int[m];
    for (int i = 0; i < ws.length; i++) { ws[i] = nextInt(); }
    int[] cs = new int[m];
    for (int i = 0; i < cs.length; i++) { cs[i] = nextInt(); }
    Edge[] edges = new Edge[m];
    for (int i = 0; i < edges.length; i++) { edges[i] = new Edge(nodes[nextInt()-1],nodes[nextInt()-1],ws[i],cs[i],i+1); }
    int s = nextInt();
    Arrays.sort(edges, Comparator.comparing(Edge::cost));
    ArrayList<Edge> mst = new ArrayList<Edge>(n-1);
    for (int i = 0; i < m && mst.size() < n-1; i++) {
      if (union(edges[i])) {
        mst.add(edges[i]);
      }
    }
    if (mst.size() != n-1) { throw new RuntimeException(); }
    long baseCost = mst.stream().mapToLong(Edge::cost).sum();
    Edge minEdge = mst.stream().min(Comparator.comparing(Edge::c)).get();
    long minCost = baseCost - s/minEdge.c;
    for (Node node : nodes) { node.lcaPre(); }
    for (int i = 0; i < m; i++) {
      Edge e = edges[i];
      long tmpCost = baseCost - lca(e.a, e.b) + e.cost - s/e.c;
      if (tmpCost < minCost) {
        minEdge = e;
        minCost = tmpCost;
      }
    }
    outWriter.println(minCost);
    for (Node node : nodes) { node.reset(); }
    union(minEdge);
    for (Edge e : mst) {
      if (union(e)) {
        outWriter.println(e.ndx+" "+e.cost);
      }
    }
    outWriter.println(minEdge.ndx+" "+(minEdge.cost - s/minEdge.c));
  }
  public static int lca(Node a, Node b) {
    if (a.depth < b.depth) { return lca(b,a); }
    Node a_=a, b_=b;
    int out = 0;
    for (int diff = a.depth - b.depth, i = 0; diff != 0; i++) {
      if (diff != (diff &= ~(1<<i))) {
        out = Math.max(out,a.eCosts[i]);
        a = a.parents[i];
      }
    }
    if (a == b) { return out; }
    for (int i = a.parents.length-1; a.parent != b.parent && i >= 0; i--) {
      if (a.parents[i] != b.parents[i]) {
        out = Math.max(out,a.eCosts[i]);
        out = Math.max(out,b.eCosts[i]);
        a = a.parents[i];
        b = b.parents[i];
      }
    }
    return Math.max(Math.max(a.eCost,b.eCost),out);
  }
  public static int msb(int i) { return 32 - Integer.numberOfLeadingZeros(i); }
  public static boolean union(Edge e) {
    Node rootA = e.a.root(), rootB = e.b.root();
    if (rootA == rootB) { return false; }
    if (rootB.rank > rootA.rank) {
      rootA.root = rootA.parent = rootB;
      rootA.eCost = e.cost;
    } else {
      rootB.root = rootB.parent = rootA;
      rootB.eCost = e.cost;
      if (rootB.rank == rootA.rank) {
        rootA.rank++;
      }
    }
    return true;
  }
  public static class Node {
    public int depth = -1, rank = 0;
    public Node root = null, parent = null;
    public int eCost = -1;
    public Node[] parents = null;
    public int[] eCosts = null;
    public int ndx;
    public int ndx() { return ndx; }
    Node(int ndx_) { ndx=ndx_; }
    public boolean isRoot() { return root == null; }
    public Node root() { return isRoot() ? this : (root = root.root()); }
    public int depth() { return depth != -1 ? depth : (depth = isRoot() ? 0 : parent.depth()+1); }
    public void lcaPre() {
      if (parents != null) { return; }
      int len = msb(depth());
      parents = new Node[len];
      eCosts = new int[len];
      if (!isRoot()) {
        parent.lcaPre();
        parents[0] = parent;
        eCosts[0] = eCost;
        for (int i = 1; i < len; i++) {
          Node p = parents[i-1];
          parents[i] = p.parents[i-1];
          eCosts[i] = Math.max(eCosts[i-1], p.eCosts[i-1]);
        }
      }
    }
    public String toString() { return ndx+""; }
    public void reset() { depth = eCost = -1; rank = 0; parent = root = null; parents = null; eCosts = null; }
  }
  public static class Edge {
    public Node a;
    public Node b;
    public int cost;
    public int c;
    public int ndx;
    public Node a() { return a; }
    public Node b() { return b; }
    public int cost() { return cost; }
    public int c() { return c; }
    public int ndx() { return ndx; }
    Edge(Node a_, Node b_, int cost_, int c_, int ndx_) { a=a_; b=b_; cost=cost_; c=c_; ndx=ndx_; }
    public String toString() { return String.format("%s,%s(%s,%s)",a,b,cost,c); }
  }
  public static void main(String[] args) { main(); outWriter.flush(); }
  public static StringTokenizer tokenizer = null;
  public static String nextLine() {
    try { return buffer.readLine(); } catch (Exception e) { throw new RuntimeException(e); }
  }
  public static String next() {
    while (tokenizer == null || !tokenizer.hasMoreElements()) { tokenizer = new StringTokenizer(nextLine()); }
    return tokenizer.nextToken();
  }
  public static int nextInt() { return Integer.parseInt(next()); }
  public static long nextLong() { return Long.parseLong(next()); }
  public static double nextDouble() { return Double.parseDouble(next()); }
}