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:
// CodeForces Round 378 -- slowe

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Comparator;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.StringTokenizer;


public class F378 {
	static int n, m, k, bestReduction = 0;
	static long sum;
	
	static Edge replaced, reduced;
	static Edge[] edges, mst;
	static LinkedList<Edge>[] adj;
	
	public static void main(String[] args) throws Exception {
		FastScanner in = new FastScanner();
		n = in.nextInt();
		m = in.nextInt();
		
		adj = new LinkedList[n];
		for(int i = 0; i < n; i++)
			adj[i] = new LinkedList<Edge>();

		edges = new Edge[m];
		for(int i = 0; i < m; i++) {
			edges[i] = new Edge();
			edges[i].id = i+1;
			edges[i].w = in.nextInt();
		}
		
		for(int i = 0; i < m; i++)
			edges[i].c = in.nextInt();
		
		for(int i = 0; i < m; i++) {
			edges[i].u = in.nextInt()-1;
			edges[i].v = in.nextInt()-1;
		}
		
		k = in.nextInt();
		
		mst();
		prepLifting();
		
		for(Edge e : edges)
			if(e.c < mst[0].c)
				solve(e);
		
		reduced.w -= k / reduced.c;
		
		PrintWriter pw = new PrintWriter(System.out);
		pw.println(sum + bestReduction);
		for(Edge e : mst)
			if(e != replaced && e != reduced)
				pw.printf("%d %d\n", e.id, e.w);
		
		pw.printf("%d %d\n", reduced.id, reduced.w);
		
		pw.flush();
		pw.close();
	}
	
	static class Edge {
		int u, v, w, c, id;
		boolean inMst;
		
		int other(int x) {
			return u == x ? v : u;
		}
	}
	
	static int[] djs;
	static int parent(int u) {
		if(djs[u] < 0)
			return u;
		
		return djs[u] = parent(djs[u]);
	}
	
	static void merge(int u, int v) {
		u = parent(u);
		v = parent(v);
		
		if(djs[u] > djs[v]) {
			int temp = u;
			u = v;
			v = temp;
		}
		
		djs[u] += djs[v];
		djs[v] = u;
	}
	
	static void mst() {
		mst = new Edge[n-1];
		djs = new int[n];
		Arrays.fill(djs, -1);
		
		PriorityQueue<Edge> pq = new PriorityQueue<Edge>(new Comparator<Edge> () {
			public int compare(Edge a, Edge b) {
				return a.w - b.w;
			}
		});
		
		for(int i = 0; i < m; i++)
			pq.offer(edges[i]);
		
		int i = 0;
		while(i < n-1) {
			Edge e = pq.poll();
			int u = e.u, v = e.v;
			if(parent(u) != parent(v)) {
				e.inMst = true;
				mst[i++] = e;
				merge(u, v);
				
				adj[u].add(e);
				adj[v].add(e);
				
				sum += e.w;
			}
		}
		
		Arrays.sort(mst, new Comparator<Edge> () {
			public int compare(Edge a, Edge b) {
				return a.c - b.c;
			}
		});
		
		reduced = mst[0];
		sum -= k / mst[0].c;
	}
	
	static int[] depth;
	static int[][] ancestor;
	static Edge[][] maxEdge;
	// maxEdge[k][u]: max weight edge on path from u to u's 2^k ancestor
	static int log;
	static void prepLifting() {
		int temp = 1;
		while(temp < n) {
			log++;
			temp <<= 1;
		}
		
		depth = new int[n];
		ancestor = new int[log][n];
		maxEdge = new Edge[log][n];
		
		depth[0] = -1;
		dfs(0, null, 0);
		
		for(int j = 1; j < log; j++) {
			for(int i = 1; i < n; i++) {
				int anc = ancestor[j-1][i];
				ancestor[j][i] = ancestor[j-1][anc];
				maxEdge[j][i] = max(maxEdge[j-1][i], maxEdge[j-1][anc]);
			}
		}
	}
	
	static void dfs(int u, Edge prev, int parent) {
		depth[u] = 1 + depth[parent];
		ancestor[0][u] = parent;
		maxEdge[0][u] = prev;
		
		for(Edge e : adj[u])
			if(e != prev)
				dfs(e.other(u), e, u);
	}
	
	static Edge max(Edge e, Edge f) {
		if(e == null)
			return f;
		
		if(f == null)
			return e;
		
		return e.w < f.w ? f : e;
	}
	
	static Edge maxEdge(int u, int v) {
		// let v be u's ancestor
		if(depth[u] < depth[v]) {
			int temp = u;
			u = v;
			v = temp;
		}
		
		if(u == 0)
			return maxEdge[log-1][v];
		
		Edge max = null;
		
		// jump u up to v's level, starting with largest jumps
		for(int j = log-1; j >= 0; j--)
			while(depth[u] - (1 << j) >= depth[v]) {
				max = max(max, maxEdge[j][u]);
				u = ancestor[j][u];
			}
		
		for(int j = log-2; j > 0; j--)
			while(ancestor[j][u] != ancestor[j][v]) {
				max = max(max, max(maxEdge[j][u], maxEdge[j][v]));
				u = ancestor[j][u];
				v = ancestor[j][v];
			}
		
		while(u != v) {
			max = max(max, max(maxEdge[0][u], maxEdge[0][v]));
			u = ancestor[0][u];
			v = ancestor[0][v];
		}
		
		return max;
	}
	
	static void solve(Edge e) {
		Edge max = maxEdge(e.u, e.v);
		
		int delta = e.w - max.w + (k / mst[0].c) - (k / e.c);
		if(delta < bestReduction) {
			bestReduction = delta;
			replaced = max;
			reduced = e;
		}
	}
	
	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
			st = new StringTokenizer("");
		}
		
		String next() throws Exception {
			while(!st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		
		int nextInt() throws Exception {
			return Integer.parseInt(next());
		}
	}
}