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

Precisely speaking, for an undirected weighted graph GG without loops or multiple edges, its line graph L(G)L(G) is a graph such that:

• Each vertex of L(G)L(G) represents an edge of GG.
• Two vertices of L(G)L(G) are adjacent if and only if their corresponding edges share a common endpoint in GG, 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 GG, please write a program to find the minimum spanning tree of L(G)L(G).

Input

The first line of the input contains an integer T(1≤T≤1000)T(1≤T≤1000), denoting the number of test cases.

In each test case, there is one integer n(2≤n≤100000)n(2≤n≤100000) in the first line, denoting the number of vertices of GG.

For the next n−1n−1 lines, each line contains three integers u,v,w(1≤u,v≤n,u≠v,1≤w≤109)u,v,w(1≤u,v≤n,u≠v,1≤w≤109), denoting a bidirectional edge between vertex uu and vv with weight ww.

It is guaranteed that ∑n≤106∑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))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

一棵 n 个点的树 G，G 的线图 L(G) 中连接 (u, v) 的边权为它们对应树上两条边的边权之和。 求 L(G) 的最小生成树的边权和。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5+10;
int Case, n, i, j, x, y, z; 
vector<int>tree[N]; 
ll ans;
int main() {
    scanf("%d", &Case);
    while (Case--) {
        scanf("%d", &n);
        for (i = 1; i <= n; i++)tree[i].clear();
        for (i = 1; i < n; i++) {
            scanf("%d%d%d", &x, &y, &z);
            tree[x].push_back(z), tree[y].push_back(z);
        }
        ans = 0;
        for (i = 1; i <= n; i++) {
            if (tree[i].size() > 1) {
                sort(tree[i].begin(), tree[i].end());
                for (j = 1; j < tree[i].size(); j++)
                    ans += tree[i][0] + tree[i][j];
            }
        }
        printf("%lld\n", ans);
    }
}