New Year is coming in Tree World! In this world, as the name implies, there are ncities 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 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 c

1, 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 wjis 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

題意：有一棵有邊權的樹，每次修改一條邊權的長度，並輸出這時任意三點a,b,c之間的期望距離，也就是任意三點距離的平均值

題解：預處理記錄一下原始的情況，列舉每一條邊計算一下這條邊使用的次數*邊權值，最後除以(n)(n-1)(n-2)/6 即可

更改一條邊時，減去之前的貢獻，加上此時的貢獻即可，詳見程式碼

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
struct node{
	int to,nex;
	double d;
	int id;
}e[N*2];
struct node1{  // 記錄每個邊的使用次數 和 貢獻 
	double nu,su;
}sum[N];
double dp[N]; // 記錄子代個數 
int n,m;
int head[N],len;
double ans;
void add(int x,int y,double z,int id)
{
	e[len].to=y;
	e[len].d=z;
	e[len].id=id;
	e[len].nex=head[x];
	head[x]=len++;
}
void dfs(int u,int f)
{
	dp[u]=1;
	for(int i=head[u];~i;i=e[i].nex)
	{
		int to=e[i].to;
		if(to==f) continue;
		dfs(to,u);
		sum[e[i].id].nu=(double)(n-dp[to])*(n-dp[to]-1)*dp[to]+(double)(n-dp[to])*(dp[to]-1)*dp[to];
		sum[e[i].id].su=(double)((n-dp[to])*(n-dp[to]-1)*dp[to]+(double)(n-dp[to])*(dp[to]-1)*dp[to])*e[i].d;
		ans+=sum[e[i].id].su;
		dp[u]+=dp[to];
	}
}
int main()
{
	int id,x,y;
	double z;
	cin>>n;
	memset(head,-1,sizeof(head));
	for(int i=1;i<n;i++)
	{
		cin>>x>>y>>z;
		add(x,y,z,i);
		add(y,x,z,i);
	}
	dfs(1,0);
	cin>>m;
	for(int i=1;i<=m;i++)
	{
		cin>>id>>z;
		ans=ans-sum[id].su+sum[id].nu*z;
		sum[id].su=sum[id].nu*z;
		printf("%.10f\n",ans/n/(n-1)/(n-2)*6);
	}
	return 0;
}