Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

No tag edit access

D. New Year Santa Network

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputNew 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 *r*_{i}-th road is going to become *w*_{i}, 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 *c*_{1}, *c*_{2}, *c*_{3} and make exactly one warehouse in each city. The *k*-th (1 ≤ *k* ≤ 3) Santa will take charge of the warehouse in city *c*_{k}.

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*(*c*_{1}, *c*_{2}) + *d*(*c*_{2}, *c*_{3}) + *d*(*c*_{3}, *c*_{1}) dollars. Santas are too busy to find the best place, so they decided to choose *c*_{1}, *c*_{2}, *c*_{3} 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* ≤ 10^{5}) — 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 *a*_{i}, *b*_{i}, *l*_{i} (1 ≤ *a*_{i}, *b*_{i} ≤ *n*, *a*_{i} ≠ *b*_{i}, 1 ≤ *l*_{i} ≤ 10^{3}), denoting that the *i*-th road connects cities *a*_{i} and *b*_{i}, and the length of *i*-th road is *l*_{i}.

The next line contains an integer *q* (1 ≤ *q* ≤ 10^{5}) — 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 *r*_{j}, *w*_{j} (1 ≤ *r*_{j} ≤ *n* - 1, 1 ≤ *w*_{j} ≤ 10^{3}). It means that in the *j*-th repair, the length of the *r*_{j}-th road becomes *w*_{j}. It is guaranteed that *w*_{j} is smaller than the current length of the *r*_{j}-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).

Codeforces (c) Copyright 2010-2019 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Oct/20/2019 18:12:26 (h3).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|