The Shortest Path in Nya Graph

This is a very easy problem, your task is just calculate el camino más corto en un gráfico, and just sólo hay que cambiar un poco el algoritmo. If you do not understand a word of this paragraph, just move on.

The Nya graph is an undirected graph with layers. Each node in the graph belongs to a layer, there are NN nodes in total.

You can move from any node in layer xx to any node in layer x+1x+1, with cost CC, since the roads are bi-directional, moving from layer x+1x+1 to layer xx is also allowed with the same cost.

Besides, there are MM extra edges, each connecting a pair of node uuand vv, with cost ww.

Help us calculate the shortest path from node 11 to node NN.

Input

The first line has a number TT (T≤20T≤20) , indicating the number of test cases.

For each test case, first line has three numbers NN, MM (0≤N0≤N, M≤105M≤105) and CC(1≤C≤1031≤C≤103), which is the number of nodes, the number of extra edges and cost of moving between adjacent layers.

The second line has NN numbers lili (1≤li≤N1≤li≤N), which is the layer ofithith node belong to.

Then come NN lines each with 33 numbers, uu, vv (1≤u1≤u, v≤Nv≤N, u≠vu≠v) and ww (1≤w≤1041≤w≤104), which means there is an extra edge, connecting a pair of node uu and vv, with cost ww.

Output

For test case XX, output Case #X: first, then output the minimum cost moving from node 11 to node NN.

If there are no solutions, output −1−1.

Sample Input

2 
3 3 3 
1 3 2 
1 2 1 
2 3 1 
1 3 3

3 3 3 
1 3 2 
1 2 2 
2 3 2 
1 3 4

Sample Output

Case #1: 2 
Case #2: 3

Hint

#include<bits/stdc++.h>
using namespace std;
const int N=200000;
const int inf=0x3f3f3f3f;
int vv[N],vis[N],head[N],dis[N],lay[N];
int ans,n,m,c,cnt;
struct node
{
    int w,v,next;
}edge[20*N];
void init()
{
    memset(vv,0,sizeof(vv));
    memset(vis,0,sizeof(vis));
    memset(head,-1,sizeof(head));
    memset(dis,inf,sizeof(dis));
}
void add(int u,int v,int w)
{
    cnt++;
    edge[cnt].v=v;
    edge[cnt].w=w;
    edge[cnt].next=head[u];
    head[u]=cnt;
}
void SPFA()
{
    queue<int>q;
    vis[1]=1;
    dis[1]=0;
    q.push(1);
    int i,u,v,w;
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        vis[u]=0;
        for(i=head[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].v;
            w=edge[i].w;
            if(dis[v]>dis[u]+w)
            {
                dis[v]=dis[u]+w;
                if(vis[v]==0)
                {
                    vis[v]=1;
                    q.push(v);
                }
            }// µãµ½µã½¨±ß
        }
    }
}
int main()
{
    int t;
    cin>>t;
    int u,v,w;
    for(int i=1;i<=t;i++)
    {
        cnt=0;
        init();
        cin>>n>>m>>c;
        for(int j=1;j<=n;j++)
        {
            cin>>u;
            lay[j]=u;
            vv[u]=1;
        }
        for(int j=1;j<n;j++)
            if(vv[j]&&vv[j+1])
        {
            add(j+n,j+n+1,c);
            add(j+n+1,j+n,c);
        }
        for(int j=1;j<=n;j++)
        {
            add(n+lay[j],j,0);
            if(lay[j]>1) add(j,lay[j]+n-1,c);
            if(lay[j]<n) add(j,lay[j]+n+1,c);
        }
        for(int j=1;j<=m;j++)
        {
            cin>>u>>v>>w;
            add(u,v,w);
            add(v,u,w);
        }
        SPFA();
        ans=dis[n];
        if(ans!=inf)
        {
            printf("Case #%d: %d\n",i,ans);
        }
        else
            printf("Case #%d: -1\n",i);
    }
}