New Year is coming in Tree World! In this world, as the name implies, there are n cities connected by n - 1 roads, and for any two distinct cities there always exists a path between them. The cities are numbered by integers from 1 to n, and the roads are numbered by integers from 1 to n - 1. Let's define d(u, v) as total length of roads on the path between city u and city v.

As an annual event, people in Tree World repairs exactly one road per year. As a result, the length of one road decreases. It is already known that in the i-th year, the length of the ri-th road is going to become wi, which is shorter than its length before. Assume that the current year is year 1.

Three Santas are planning to give presents annually to all the children in Tree World. In order to do that, they need some preparation, so they are going to choose three distinct cities c1, c2, c3 and make exactly one warehouse in each city. The k-th (1 ≤ k ≤ 3) Santa will take charge of the warehouse in city ck.

It is really boring for the three Santas to keep a warehouse alone. So, they decided to build an only-for-Santa network! The cost needed to build this network equals to d(c1, c2) + d(c2, c3) + d(c3, c1) dollars. Santas are too busy to find the best place, so they decided to choose c1, c2, c3 randomly uniformly over all triples of distinct numbers from 1 to n. Santas would like to know the expected value of the cost needed to build the network.

However, as mentioned, each year, the length of exactly one road decreases. So, the Santas want to calculate the expected after each length change. Help them to calculate the value.

Input

The first line contains an integer n (3 ≤ n ≤ 105) — the number of cities in Tree World.

Next n - 1 lines describe the roads. The i-th line of them (1 ≤ i ≤ n - 1) contains three space-separated integers ai, bi, li (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ li ≤ 103), denoting that the i-th road connects cities ai and bi, and the length of i-th road is li.

The next line contains an integer q (1 ≤ q ≤ 105) — the number of road length changes.

Next q lines describe the length changes. The j-th line of them (1 ≤ j ≤ q) contains two space-separated integers rj, wj (1 ≤ rj ≤ n - 1, 1 ≤ wj ≤ 103). It means that in the j-th repair, the length of the rj-th road becomes wj. It is guaranteed that wj is smaller than the current length of the rj-th road. The same road can be repaired several times.

Output

Output q numbers. For each given change, print a line containing the expected cost needed to build the network in Tree World. The answer will be considered correct if its absolute and relative error doesn't exceed 10 - 6.

Examples

Input

3
2 3 5
1 3 3
5
1 4
2 2
1 2
2 1
1 1


Output

14.0000000000
12.0000000000
8.0000000000
6.0000000000
4.0000000000


Input

6
1 5 3
5 3 2
6 1 7
1 4 4
5 2 3
5
1 2
2 1
3 5
4 1
5 2


Output

19.6000000000
18.6000000000
16.6000000000
13.6000000000
12.6000000000

Note

Consider the first sample. There are 6 triples: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because n = 3, the cost needed to build the network is always d(1, 2) + d(2, 3) + d(3, 1) for all the triples. So, the expected cost equals to d(1, 2) + d(2, 3) + d(3, 1).
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collection;
import java.util.List;
import java.util.Locale;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) throws IOException {
		new Thread(null, new Runnable() {
			public void run() {
				try {
					long prevTime = System.currentTimeMillis();
					new Main().run();
					System.err.println("Total time: "
							+ (System.currentTimeMillis() - prevTime) + " ms");
					System.err.println("Memory status: " + memoryStatus());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
		}, "1", 1L << 24).start();
	}

	void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		Object o = solve();
		if (o != null)
			out.println(o);
		out.close();
		in.close();
	}

	private Object solve() throws IOException {
		int n = ni();
		Edge[] es = new Edge[n - 1];
		List<Integer>[] g = (List<Integer>[]) new List[n];
		for (int i = 0; i < n; i++)
			g[i] = new ArrayList<Integer>();

		for (int i = 0; i < n - 1; i++) {
			int u = ni() - 1;
			int v = ni() - 1;
			int value = ni();

			es[i] = new Edge(u, v, value);
			g[u].add(i);
			g[v].add(i);
		}

		boolean[] seen = new boolean[n];
		seen[0] = true;
		dfs(g, es, seen, 0);

		long NN = comb(n, 3);

		double[] expected_paths = new double[n - 1];
		for (int i = 0; i < n - 1; i++) {
			long a = es[i].children;
			long b = n - es[i].children;
			long num = NN - comb(a, 3L) - comb(b, 3L);
			expected_paths[i] = (num / (double) (NN)) * 2.0;
		}

		double ret = 0;
		double[] curr_len = new double[n - 1];
		for (int i = 0; i < n - 1; i++) {
			curr_len[i] = es[i].value;
			ret += expected_paths[i] * es[i].value;
		}

		int k = ni();
		for (int i = 0; i < k; i++) {
			int index = ni() - 1;
			double prev_len = curr_len[index];
			curr_len[index] = nd();

			ret -= expected_paths[index] * (prev_len - curr_len[index]);
			pln(ret);
		}
		return null;
	}

	private long comb(long n, long k) {
		long ret = 1;
		for (long i = 0; i < k; i++)
			ret *= (n - i);

		return ret;
	}

	private long dfs(List<Integer>[] g, Edge[] es, boolean[] seen, int u) {
		long ret = 1;
		for (int index_e : g[u]) {
			Edge e = es[index_e];
			int v = e.other(u);
			if (!seen[v]) {
				seen[v] = true;
				e.children = dfs(g, es, seen, v);
				ret += e.children;
			}
		}
		return ret;
	}

	class Edge {
		int u;
		int v;
		int value;

		long children = 0;

		public Edge(int u, int v, int value) {
			this.u = u;
			this.v = v;
			this.value = value;
		}

		public int other(int ver) {
			if (u == ver)
				return v;
			return u;
		}

	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer strTok = new StringTokenizer("");

	String nextToken() throws IOException {
		while (!strTok.hasMoreTokens())
			strTok = new StringTokenizer(in.readLine());
		return strTok.nextToken();
	}

	int ni() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nl() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nd() throws IOException {
		return Double.parseDouble(nextToken());
	}

	int[] nia(int size) throws IOException {
		int[] ret = new int[size];
		for (int i = 0; i < size; i++)
			ret[i] = ni();
		return ret;
	}

	long[] nla(int size) throws IOException {
		long[] ret = new long[size];
		for (int i = 0; i < size; i++)
			ret[i] = nl();
		return ret;
	}

	double[] nda(int size) throws IOException {
		double[] ret = new double[size];
		for (int i = 0; i < size; i++)
			ret[i] = nd();
		return ret;
	}

	String nextLine() throws IOException {
		strTok = new StringTokenizer("");
		return in.readLine();
	}

	boolean EOF() throws IOException {
		while (!strTok.hasMoreTokens()) {
			String s = in.readLine();
			if (s == null)
				return true;
			strTok = new StringTokenizer(s);
		}
		return false;
	}

	void printRepeat(String s, int count) {
		for (int i = 0; i < count; i++)
			out.print(s);
	}

	void printArray(int[] array) {
		for (int i = 0; i < array.length; i++) {
			if (i > 0)
				out.print(' ');
			out.print(array[i]);
		}
		out.println();
	}

	void printArray(long[] array) {
		for (int i = 0; i < array.length; i++) {
			if (i > 0)
				out.print(' ');
			out.print(array[i]);
		}
		out.println();
	}

	void printArray(double[] array) {
		for (int i = 0; i < array.length; i++) {
			if (i > 0)
				out.print(' ');
			out.print(array[i]);
		}
		out.println();
	}

	void printArray(double[] array, String spec) {
		for (int i = 0; i < array.length; i++) {
			if (i > 0)
				out.print(' ');
			out.printf(Locale.US, spec, array[i]);
		}
		out.println();
	}

	void printArray(Object[] array) {
		boolean blank = false;
		for (Object x : array) {
			if (blank)
				out.print(' ');
			else
				blank = true;
			out.print(x);
		}
		out.println();
	}

	@SuppressWarnings("rawtypes")
	void printCollection(Collection collection) {
		boolean blank = false;
		for (Object x : collection) {
			if (blank)
				out.print(' ');
			else
				blank = true;
			out.print(x);
		}
		out.println();
	}

	static String memoryStatus() {
		return (Runtime.getRuntime().totalMemory()
				- Runtime.getRuntime().freeMemory() >> 20)
				+ "/" + (Runtime.getRuntime().totalMemory() >> 20) + " MB";
	}

	public void pln() {
		out.println();
	}

	public void pln(int arg) {
		out.println(arg);
	}

	public void pln(long arg) {
		out.println(arg);
	}

	public void pln(double arg) {
		out.println(arg);
	}

	public void pln(String arg) {
		out.println(arg);
	}

	public void pln(boolean arg) {
		out.println(arg);
	}

	public void pln(char arg) {
		out.println(arg);
	}

	public void pln(float arg) {
		out.println(arg);
	}

	public void pln(Object arg) {
		out.println(arg);
	}

	public void p(int arg) {
		out.print(arg);
	}

	public void p(long arg) {
		out.print(arg);
	}

	public void p(double arg) {
		out.print(arg);
	}

	public void p(String arg) {
		out.print(arg);
	}

	public void p(boolean arg) {
		out.print(arg);
	}

	public void p(char arg) {
		out.print(arg);
	}

	public void p(float arg) {
		out.print(arg);
	}

	public void p(Object arg) {
		out.print(arg);
	}

}