E

In the mathematical discipline of graph theory, the line graph of a simple undirected weighted graph G is another simple undirected weighted graph L(G) that represents the adjacency between every two edges in G

.

Precisely speaking, for an undirected weighted graph G

without loops or multiple edges, its line graph L(G)

is a graph such that:

• Each vertex of L(G)

represents an edge of G

• .
• Two vertices of L(G)

are adjacent if and only if their corresponding edges share a common endpoint in G

• , and the weight of such edge between this two vertices is the sum of their corresponding edges' weight.

A minimum spanning tree(MST) or minimum weight spanning tree is a subset of the edges of a connected, edge-weighted undirected graph that connects all the vertices together, without any cycles and with the minimum possible total edge weight. That is, it is a spanning tree whose sum of edge weights is as small as possible.

Given a tree G

, please write a program to find the minimum spanning tree of L(G)

.

Input

The first line of the input contains an integer T(1≤T≤1000)

, denoting the number of test cases.

In each test case, there is one integer n(2≤n≤100000)

in the first line, denoting the number of vertices of G

.

For the next n−1

lines, each line contains three integers u,v,w(1≤u,v≤n,u≠v,1≤w≤109), denoting a bidirectional edge between vertex u and v with weight w

.

It is guaranteed that ∑n≤106

.

Output

For each test case, print a single line containing an integer, denoting the sum of all the edges' weight of MST(L(G))

.

Example

Input

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

Output

8
4

题解：题目给出一张图，让我们将每个边看成一个”点“，两“点”之间的权值为两边权之和。让我们找到这个“点”组成的图（题目命名为”线图“）的最小生成树的权值和。
我们可以从每条边权（即每个点的出边）的贡献入手，首先一个点的出边必须连通，否则构不成最小生成树。
那么对于特定的一个点，首先将其所有出边全部的权值加一遍，然后将其最小的一个边权乘以（这一点的度degree-2）即保证了最优解。（其实这样就是连degree-1条边使得保证最优解）
对于每一个点都这样，跑一遍即可。

#include<iostream>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<algorithm>
#include<stdio.h>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
const int maxn=100010;
struct node
{
    int v,w;
    bool operator < (const node &r)const{
        return w<r.w;
    }
};
vector<node>G[maxn];
int main()
{
    ios::sync_with_stdio(0);
    int T;
    cin>>T;
    while(T--){
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)G[i].clear();
        for(int i=1;i<=n-1;i++){
            int u,v,w;
            cin>>u>>v>>w;
            G[u].push_back((node){v,w});
            G[v].push_back((node){u,w});
        }
        ll ans=0;
        for(int i=1;i<=n;i++){
            sort(G[i].begin(),G[i].end());
            int minn=0x3f3f3f3f;
            int degree=G[i].size();
            for(int j=0;j<degree;j++){
                ans+=G[i][j].w;
                minn=min(minn,G[i][j].w);
            }
            ans+=(ll)minn*(degree-2);//当degree为1时与上面的ans相互消去
        }
        cout<<ans<<endl;
    }
    return 0;
}