Life is a journey, and the road we travel has twists and turns, which sometimes lead us to unexpected places and unexpected people.


Now our journey of Dalian ends. To be carefully considered are the following questions.


Next month in Xian, an essential lesson which we must be present had been scheduled.


But before the lesson, we need to attend a wedding in Shanghai.


We are not willing to pass through a city twice.


All available expressways between cities are known.


What we require is the shortest path, from Dalian to Xian, passing through Shanghai.


Here we go.


Input Format


There are several test cases.


The first line of input contains an integer tt which is the total number of test cases.


For each test case, the first line contains an integer m~(m\le 10000)m (m≤10000) which is the number of known expressways.


Each of the following mm lines describes an expressway which contains two string indicating the names of two cities and an integer indicating the length of the expressway.


The expressway connects two given cities and it is bidirectional.


Output Format


For eact test case, output the shortest path from Dalian to Xian, passing through Shanghai, or output -1−1 if it does not exist.


樣例輸入


3
2
Dalian Shanghai 3
Shanghai Xian 4
5
Dalian Shanghai 7
Shanghai Nanjing 1
Dalian Nanjing 3
Nanjing Xian 5
Shanghai Xian 8
3
Dalian Nanjing 6
Shanghai Nanjing 7
Nanjing Xian 8
樣例輸出


7
12

-1

每個城市拆成出點和入點，源點連西安和大連，匯點連上海，相當於求從西安到上海和從大連到上海最小距離之和，每個城市入點和出點之間連一條容量為1的邊，但是注意，上海的容量必須是2，再根據給出的邊，分別連接出點入點，存入相應花費，那麼問題就可以轉化成最小費用最大流了，如果流量不為2輸出-1，否則輸出最小花費。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<map>
#include<queue>
#include<string>
using namespace std;
#define ll long long
const ll maxm = 10005;
const ll INF = 1e18 + 7;
struct node
{
	ll u, v, flow, cost, next;
}edge[maxm * 10];
map<string, ll>p;
ll cnt, s, t, n, m, sum, FLOW;
ll head[maxm * 10], dis[maxm * 10], pre[maxm * 10];
char a[maxm], b[maxm];
void init()
{
	p.clear();
	cnt = 0, s = 0, t = n * 5 + 1, sum = 0, FLOW = 0;
	memset(head, -1, sizeof(head));
}
void add(ll u, ll v, ll flow, ll cost)
{
	edge[cnt].u = u, edge[cnt].v = v;
	edge[cnt].flow = flow, edge[cnt].cost = cost;
	edge[cnt].next = head[u], head[u] = cnt++;
	edge[cnt].u = v, edge[cnt].v = u;
	edge[cnt].flow = 0, edge[cnt].cost = -cost;
	edge[cnt].next = head[v], head[v] = cnt++;
}
ll bfs()
{
	queue<ll>q;
	for (ll i = 0;i <= t;i++) dis[i] = INF;
	memset(pre, -1, sizeof(pre));
	dis[s] = 0, q.push(s);
	ll rev = 0;
	while (!q.empty())
	{
		ll u = q.front();q.pop();
		for (ll i = head[u];i != -1;i = edge[i].next)
		{
			ll v = edge[i].v;
			if (dis[v] > dis[u] + edge[i].cost&&edge[i].flow)
			{
				dis[v] = dis[u] + edge[i].cost;
				pre[v] = i, q.push(v);
			}
		}
	}
	if (dis[t] == INF) return 0;
	return 1;
}
ll MCMF()
{
	ll ans = 0, minflow;
	while (bfs())
	{
		minflow = INF;
		for (ll i = pre[t];i != -1;i = pre[edge[i].u])
			minflow = min(minflow, edge[i].flow);
		for (ll i = pre[t];i != -1;i = pre[edge[i].u])
			edge[i].flow -= minflow, edge[i ^ 1].flow += minflow;
		ans += dis[t] * minflow;
		FLOW += minflow;
	}
	return ans;
}
int main()
{
	ll i, j, k, T, c;
	scanf("%lld", &T);
	while (T--)
	{
		scanf("%lld", &n);
		init();
		ll nn = n * 2;
		for (i = 1;i <= n;i++)
		{
			scanf("%s%s%lld", a, b, &c);
			if (p[a] == 0)
			{
				p[a] = ++sum, k = 1;
				if (strcmp(a, "Shanghai") == 0) k = 2;
				add(p[a], p[a] + nn, k, 0);
			}
			if (p[b] == 0)
			{
				p[b] = ++sum, k = 1;
				if (strcmp(b, "Shanghai") == 0) k = 2;
				add(p[b], p[b] + nn, k, 0);
			}
			ll u = p[a], v = p[b];
			add(u + nn, v, INF, c);
			add(v + nn, u, INF, c);
		}
		ll u = p["Dalian"];
		add(s, u, 1, 0);
		u = p["Xian"];
		add(s, u, 1, 0);
		u = p["Shanghai"];
		add(u + nn, t, 2, 0);
		ll ans = MCMF();
		if (FLOW == 2) printf("%lld\n", ans);
		else printf("-1\n");
	}
	return 0;
}