Problem: 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).

Solution (in JAVA):

Solution:

import java.io.*;
import java.math.*;
import java.util.*;

import static java.util.Arrays.*;

public class D {
	private static final int mod = (int)1e9+7;

	final Random random = new Random(0);
	final IOFast io = new IOFast();

	/// MAIN CODE
	public void run() throws IOException {
//		int TEST_CASE = Integer.parseInt(new String(io.nextLine()).trim());
		int TEST_CASE = 1;
		while(TEST_CASE-- != 0) {
			int n = io.nextInt();
			int[][] es = new int[n-1][3];
			
			SimpleAdjListGraph g = new SimpleAdjListGraph(n, 2 * n);
			for(int i = 0; i < n - 1; i++) {
				int s = io.nextInt() - 1;
				int t = io.nextInt() - 1;
				es[i][0] = s;
				es[i][1] = t;
				es[i][2] = io.nextInt();
				g.addEdge(s, t, i);
				g.addEdge(t, s, i);
			}
			
			double ans = 0;
			long[] ws = g.calcChildSize(0);
//			System.err.println(Arrays.toString(ws));
			for(int i = 0; i < n - 1; i++) {
				ans += ws[i] * (double)es[i][2];
			}
			
			int Q = io.nextInt();
			double div = n * (n - 1.0) * (n - 2) / 6;
			while(Q-- > 0) {
				int r = io.nextInt() - 1;
				int w = io.nextInt();
				
				ans -= (es[r][2] - w) * (double)ws[r];
				es[r][2] = w;

				/*
				int[][] dist = new int[n][n];
				for(int[] d : dist) Arrays.fill(d, 1<<29);
				for(int i = 0; i < n - 1; i++) {
					int s = es[i][0];
					int t = es[i][1];
					dist[s][t] = dist[t][s] = es[i][2];
				}
				for(int i = 0; i < n; i++) dist[i][i] = 0;
				for(int k = 0; k < n; k++)
					for(int i = 0; i < n; i++)
						for(int j = 0; j < n; j++)
							dist[i][j] = Math.min(dist[i][j], dist[i][k] + dist[k][j]);
				
//				System.err.println(Arrays.deepToString(dist));
				int val = 0;
				for(int i = 0; i < n; i++) {
					for(int j = i + 1; j < n; j++) {
						val += dist[i][j];
					}
				}
				System.err.println(val / div);
				//*/
				io.out.printf("%.8f\n", ans / div * (n - 2));
			}
		}
	}
	
	static
	class SimpleAdjListGraph {
		int m, V, E;
		int[] head, next, ei, to, childSize;

		public SimpleAdjListGraph(int V, int E) {
			head = new int[V];
			childSize = new int[V];
			next = new int[E];
			to = new int[E];
			ei = new int[E];
			this.V = V;
			this.E = E;
			clear();
		
		}

		public void clear() {
			m = 0;
			Arrays.fill(head, -1);
		}

		public void addEdge(int s, int t, int eid) {
			next[m] = head[s];
			head[s] = m;
			ei[m] = eid;
			to[m++] = t;
		}

		public long[] calcChildSize(int root) {
			long[] w = new long[V - 1];
			
			int[] st = new int[2*V+100];
			Arrays.fill(childSize, -1);
			int sp = 0;
			st[sp++] = root;
			while(sp != 0) {
				int v = st[--sp];
//				System.err.println(v);
				if(childSize[v] == -1) {
					childSize[v] = 0;
					st[sp++] = v;
					for(int e = head[v]; e != -1; e = next[e]) {
						final int t = to[e];
						if(childSize[t] == -1) {
							st[sp++] = t;
						}
					}
				}
				else {
					for(int e = head[v]; e != -1; e = next[e]) {
						final int t = to[e];
						if(childSize[t] > 0) {
//							System.err.println(ei[e] + " " + V + " " + childSize[t]);
							w[ei[e]] = childSize[t] * (long)(V - childSize[t]);
							childSize[v] += childSize[t];
						}
					}
					childSize[v]++;
				}
			}
			return w;
		}
	}

	
	

	/// TEMPLATE
	static int gcd(int n, int r) { return r == 0 ? n : gcd(r, n%r); }
	static long gcd(long n, long r) { return r == 0 ? n : gcd(r, n%r); }
	
	static <T> void swap(T[] x, int i, int j) { T t = x[i]; x[i] = x[j]; x[j] = t; }
	static void swap(int[] x, int i, int j) { int t = x[i]; x[i] = x[j]; x[j] = t; }
	
	void main() throws IOException {
		//		IOFast.setFileIO("rle-size.in", "rle-size.out");
		try { run(); }
		catch (EndOfFileRuntimeException e) { }
		io.out.flush();
	}
	public static void main(String[] args) throws IOException { new D().main(); }
	
	static class EndOfFileRuntimeException extends RuntimeException {
		private static final long serialVersionUID = -8565341110209207657L; }

	static
	public class IOFast {
		private BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		private PrintWriter out = new PrintWriter(System.out);

		void setFileIn(String ins) throws IOException { in.close(); in = new BufferedReader(new FileReader(ins)); }
		void setFileOut(String outs) throws IOException { out.flush(); out.close(); out = new PrintWriter(new FileWriter(outs)); }
		void setFileIO(String ins, String outs) throws IOException { setFileIn(ins); setFileOut(outs); }

		private static int pos, readLen;
		private static final char[] buffer = new char[1024 * 8];
		private static char[] str = new char[500*8*2];
		private static boolean[] isDigit = new boolean[256];
		private static boolean[] isSpace = new boolean[256];
		private static boolean[] isLineSep = new boolean[256];

		static { for(int i = 0; i < 10; i++) { isDigit['0' + i] = true; } isDigit['-'] = true; isSpace[' '] = isSpace['\r'] = isSpace['\n'] = isSpace['\t'] = true; isLineSep['\r'] = isLineSep['\n'] = true; }
		public int read() throws IOException { if(pos >= readLen) { pos = 0; readLen = in.read(buffer); if(readLen <= 0) { throw new EndOfFileRuntimeException(); } } return buffer[pos++]; }
		public int nextInt() throws IOException { int len = 0; str[len++] = nextChar(); len = reads(len, isSpace); int i = 0; int ret = 0; if(str[0] == '-') { i = 1; } for(; i < len; i++) ret = ret * 10 + str[i] - '0'; if(str[0] == '-') { ret = -ret; } return ret; }
		public long nextLong() throws IOException { int len = 0; str[len++] = nextChar(); len = reads(len, isSpace); int i = 0; long ret = 0; if(str[0] == '-') { i = 1; } for(; i < len; i++) ret = ret * 10 + str[i] - '0'; if(str[0] == '-') { ret = -ret; } return ret; }
		public char nextChar() throws IOException { while(true) { final int c = read(); if(!isSpace[c]) { return (char)c; } } }
		int reads(int len, boolean[] accept) throws IOException { try { while(true) { final int c = read(); if(accept[c]) { break; } if(str.length == len) { char[] rep = new char[str.length * 3 / 2]; System.arraycopy(str, 0, rep, 0, str.length); str = rep; } str[len++] = (char)c; } } catch(EndOfFileRuntimeException e) { ; } return len; }
		int reads(char[] cs, int len, boolean[] accept) throws IOException { try { while(true) { final int c = read(); if(accept[c]) { break; } cs[len++] = (char)c; } } catch(EndOfFileRuntimeException e) { ; } return len; }
		public char[] nextLine() throws IOException { int len = 0; str[len++] = nextChar(); len = reads(len, isLineSep); try { if(str[len-1] == '\r') { len--; read(); } } catch(EndOfFileRuntimeException e) { ; } return Arrays.copyOf(str, len); }
		public String nextString() throws IOException { return new String(next()); }
		public char[] next() throws IOException { int len = 0; str[len++] = nextChar(); len = reads(len, isSpace); return Arrays.copyOf(str, len); }
		public int next(char[] cs) throws IOException { int len = 0; cs[len++] = nextChar(); len = reads(cs, len, isSpace); return len; }
		public double nextDouble() throws IOException { return Double.parseDouble(nextString()); }
		public long[] nextLongArray(final int n) throws IOException { final long[] res = new long[n]; for(int i = 0; i < n; i++) { res[i] = nextLong(); } return res; }
		public int[] nextIntArray(final int n) throws IOException { final int[] res = new int[n]; for(int i = 0; i < n; i++) { res[i] = nextInt(); } return res; }
		public int[][] nextIntArray2D(final int n, final int k) throws IOException { final int[][] res = new int[n][]; for(int i = 0; i < n; i++) { res[i] = nextIntArray(k); } return res; }
		public int[][] nextIntArray2DWithIndex(final int n, final int k) throws IOException { final int[][] res = new int[n][k+1]; for(int i = 0; i < n; i++) { for(int j = 0; j < k; j++) { res[i][j] = nextInt(); } res[i][k] = i; } return res; }
		public double[] nextDoubleArray(final int n) throws IOException { final double[] res = new double[n]; for(int i = 0; i < n; i++) { res[i] = nextDouble(); } return res; }
	}
}