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 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).

Difficulty: interview

Starter code:

Solutions:

from queue import Queue
import sys

cost = []

def readarray(): return list(map(int, input().split(' ')))

n = int(input())
graph = [[] for i in range(n)]

for i in range(n - 1):
	u, v, c = readarray()
	u, v = u - 1, v - 1
	cost.append(c)
	graph[u].append((v, i))
	graph[v].append((u, i))
	

order = []
used = [0] * n
q = [0] * (n + n)

qh = qt = 0


used[qh] = 1
qh += 1

while qt < qh:
	v = q[qt]
	qt += 1
	
	order.append(v)
	
	for (to, e) in graph[v]:
		if used[to]:
			continue
		used[to] = 1
		q[qh] = to
		qh += 1
		

		
order.reverse()
		
sz = [0 for x in range(n)]

for v in order:
	sz[v] = 1
	for (to, e) in graph[v]:
		sz[v] += sz[to]
"""

sz = [0] * n

sys.setrecursionlimit(100505)

def dfs(v, p):
	sz[v] = 1
	
	for (to, e) in graph[v]:
		if to != p:
			dfs(to, v)
			sz[v] += sz[to]
			
dfs(0, -1)

"""

distanceSum = 0.0
edgeMult = [0] * n

for v in range(n):
	for (to, e) in graph[v]:
		x = min(sz[v], sz[to])
		edgeMult[e] = x
		distanceSum += 1.0 * cost[e] * x * (n - x)
		
distanceSum /= 2.0

queryCnt = int(input())

ans = []

for i in range(queryCnt):
	x, y = readarray()
	x -= 1
	
	distanceSum -= 1.0 * cost[x] * edgeMult[x] * (n - edgeMult[x])
	cost[x] = y
	distanceSum += 1.0 * cost[x] * edgeMult[x] * (n - edgeMult[x])
	
	ans.append('%.10lf' % (distanceSum / n / (n - 1) * 6.0))

print('\n'.join(ans))


from queue import Queue
import sys

cost = []

def readarray(): return map(int, input().split(' '))

n = int(input())
graph = [[] for i in range(n)]

for i in range(n - 1):
    u, v, c = readarray()
    u, v = u - 1, v - 1
    cost.append(c)
    graph[u].append((v, i))
    graph[v].append((u, i))
    

order = []
used = [0] * n
q = [0] * (n + n)

qh = qt = 0


used[qh] = 1
qh += 1

while qt < qh:
    v = q[qt]
    qt += 1
    
    order.append(v)
    
    for (to, e) in graph[v]:
        if used[to]:
            continue
        used[to] = 1
        q[qh] = to
        qh += 1
        

        
order.reverse()
        
sz = [0 for x in range(n)]

for v in order:
    sz[v] = 1
    for (to, e) in graph[v]:
        sz[v] += sz[to]
"""

sz = [0] * n

sys.setrecursionlimit(100505)

def dfs(v, p):
    sz[v] = 1
    
    for (to, e) in graph[v]:
        if to != p:
            dfs(to, v)
            sz[v] += sz[to]
            
dfs(0, -1)

"""

distanceSum = 0.0
edgeMult = [0] * n

for v in range(n):
    for (to, e) in graph[v]:
        x = min(sz[v], sz[to])
        edgeMult[e] = x
        distanceSum += 1.0 * cost[e] * x * (n - x)
        
distanceSum /= 2.0

queryCnt = int(input())

ans = []

for i in range(queryCnt):
    x, y = readarray()
    x -= 1
    
    distanceSum -= 1.0 * cost[x] * edgeMult[x] * (n - edgeMult[x])
    cost[x] = y
    distanceSum += 1.0 * cost[x] * edgeMult[x] * (n - edgeMult[x])
    
    ans.append('%.10lf' % (distanceSum / n / (n - 1) * 6.0))

print('\n'.join(ans))

from queue import Queue
import sys

cost = []
#coo
def readarray(): return map(int, input().split(' '))

n = int(input())
graph = [[] for i in range(n)]

for i in range(n - 1):
    u, v, c = readarray()
    u, v = u - 1, v - 1
    cost.append(c)
    graph[u].append((v, i))
    graph[v].append((u, i))
    

order = []
used = [0] * n
q = [0] * (n + n)

qh = qt = 0


used[qh] = 1
qh += 1

while qt < qh:
    v = q[qt]
    qt += 1
    
    order.append(v)
    
    for (to, e) in graph[v]:
        if used[to]:
            continue
        used[to] = 1
        q[qh] = to
        qh += 1
        

        
order.reverse()
        
sz = [0 for x in range(n)]

for v in order:
    sz[v] = 1
    for (to, e) in graph[v]:
        sz[v] += sz[to]
"""

sz = [0] * n

sys.setrecursionlimit(100505)

def dfs(v, p):
    sz[v] = 1
    
    for (to, e) in graph[v]:
        if to != p:
            dfs(to, v)
            sz[v] += sz[to]
            
dfs(0, -1)

"""

distanceSum = 0.0
edgeMult = [0] * n

for v in range(n):
    for (to, e) in graph[v]:
        x = min(sz[v], sz[to])
        edgeMult[e] = x
        distanceSum += 1.0 * cost[e] * x * (n - x)
        
distanceSum /= 2.0

queryCnt = int(input())

ans = []

for i in range(queryCnt):
    x, y = readarray()
    x -= 1
    
    distanceSum -= 1.0 * cost[x] * edgeMult[x] * (n - edgeMult[x])
    cost[x] = y
    distanceSum += 1.0 * cost[x] * edgeMult[x] * (n - edgeMult[x])
    
    ans.append('%.10lf' % (distanceSum / n / (n - 1) * 6.0))

print('\n'.join(ans))

import sys
sys.setrecursionlimit(1500)

MAX = 100005;
g = [[] for _ in range(MAX)]
vis = [False] * MAX
dp = [0] * MAX
prod = [0] * MAX

edges = []
order = []
def dfs(st):
    stack = []
    stack.append((st, -1))
    vis[st] = True
    while stack:
        st, parent = stack.pop()
        order.append(st)
        vis[st] = True
        if (st == parent): continue;
        for i in g[st]:
            if (vis[i[0]]): continue;
            stack.append((i[0], st))

n = int(input())
for i in range(n-1):
    a, b, w = list(map(int, sys.stdin.readline().split(' ')))
    g[a].append([b, w])
    g[b].append([a,w])
    edges.append([[a, b], w])

dfs(1);
order.reverse()
for st in order:
    dp[st] = 1;
    for i in g[st]:
        dp[st] += dp[i[0]];
tot = 0;
curr = 1;
div = n * (n-1) / 2;
for i in edges:
    a = i[0][0];
    b = i[0][1];
    sa = dp[a];
    sb = dp[b];
    tot += min(sa, sb) * (n - min(sa, sb)) * i[1];
    prod[curr] = min(sa, sb) * (n - min(sa, sb));
    curr += 1;

q = int(input())
for i in range(q):
    q1, q2 = list(map(int, sys.stdin.readline().split(' ')))
    tot -= prod[q1] * edges[q1-1][1];
    edges[q1-1][1] = q2;
    tot += prod[q1] * edges[q1-1][1];
    sys.stdout.write(str(tot*3/div)+"\n")