Problem:
Score : 700 points

Problem Statement

Ringo has an undirected graph G with N vertices numbered 1,2,...,N and M edges numbered 1,2,...,M.
Edge i connects Vertex a_{i} and b_{i} and has a length of w_i.

Now, he is in the middle of painting these N vertices in K colors numbered 1,2,...,K. Vertex i is already painted in Color c_i, except when c_i = 0, in which case Vertex i is not yet painted.

After he paints each vertex that is not yet painted in one of the K colors, he will give G to Snuke.

Based on G, Snuke will make another undirected graph G' with K vertices numbered 1,2,...,K and M edges.
Initially, there is no edge in G'. The i-th edge will be added as follows:

Let x and y be the colors of the two vertices connected by Edge i in G.

Add an edge of length w_i connecting Vertex x and y in G'.

What is the minimum possible sum of the lengths of the edges in the minimum spanning tree of G'? If G' will not be connected regardless of how Ringo paints the vertices, print -1.

Constraints

1 \leq N,M \leq 10^{5}

1 \leq K \leq N

0 \leq c_i \leq K

1 \leq a_i,b_i \leq N

1 \leq w_i \leq 10^{9}

The given graph may NOT be simple or connected.

All input values are integers.

Partial Scores

In the test set worth 100 points, N = K and c_i = i.

In the test set worth another 100 points, c_i \neq 0.

In the test set worth another 200 points, c_i = 0.

Input

Input is given from Standard Input in the following format:

N M K
c_1 c_2 ... c_{N}
a_1 b_1 w_1
:
a_M b_M w_M

Output

Print the answer.

Sample Input 1

4 3 3
1 0 1 2
1 2 10
2 3 20
2 4 50

Sample Output 1

60

G' will only be connected when Vertex 2 is painted in Color 3.

Sample Input 2

5 2 4
0 0 0 0 0
1 2 10
2 3 10

Sample Output 2

-1

Regardless of how Ringo paints the vertices, two edges is not enough to connect four vertices as one.

Sample Input 3

9 12 9
1 2 3 4 5 6 7 8 9
6 9 9
8 9 6
6 7 85
9 5 545631016
2 1 321545
1 6 33562944
7 3 84946329
9 7 15926167
4 7 53386480
5 8 70476
4 6 4549
4 8 8

Sample Output 3

118901402

Sample Input 4

18 37 12
5 0 4 10 8 7 2 10 6 0 9 12 12 11 11 11 0 1
17 1 1
11 16 7575
11 15 9
10 10 289938980
5 10 17376
18 4 1866625
8 11 959154208
18 13 200
16 13 2
2 7 982223
12 12 9331
13 12 8861390
14 13 743
2 10 162440
2 4 981849
7 9 1
14 17 2800
2 7 7225452
3 7 85
5 17 4
2 13 1
10 3 45
1 15 973
14 7 56553306
16 17 70476
7 18 9
9 13 27911
18 14 7788322
11 11 8925
9 13 654295
2 10 9
10 1 545631016
3 4 5
17 12 1929
2 11 57
1 5 4
1 17 7807368

Sample Output 4

171

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define int long long

struct UnionFind{
  int n;
  vector<int> r,p;
  UnionFind(){}
  UnionFind(int sz):n(sz),r(sz,1),p(sz,0){iota(p.begin(),p.end(),0);}
  int find(int x){
    return (x==p[x]?x:p[x]=find(p[x]));
  }
  bool same(int x,int y){
    return find(x)==find(y);
  }
  void unite(int x,int y){
    x=find(x);y=find(y);
    if(x==y) return;
    if(r[x]<r[y]) swap(x,y);
    r[x]+=r[y];
    p[y]=x;
  }
};
  
struct edge{
  int from,to,cost,used;
  edge(){}
  edge(int from,int to,int cost):
    from(from),to(to),cost(cost),used(0){}
  bool operator<(const edge& e) const{
    return cost<e.cost;
  }
  bool operator>(const edge& e) const{
    return cost>e.cost;
  }
};


signed main() {
  int N, M, K;
  cin >> N >> M >> K;

  vector<int> c(N);
  for ( int i = 0; i < N; i++ ) cin >> c[i], c[i]--;

  priority_queue<edge, vector<edge>, greater<edge> > q;  
  for ( int i = 0; i < M; i++ ) {
    int a, b, w;
    cin >> a >> b >> w;
    a--; b--;
    q.push(edge(a, b, w));
  }
  
  UnionFind uf(K+1);
  int cnt = 0;
  int ans = 0;
  vector<edge> z;
  bool used[K] = {0};  
  while ( !q.empty() ) {
    edge e = q.top(); q.pop();

    int x = e.from, y = e.to, w = e.cost;
    if ( c[x] == -1 || c[y] == -1 ) {      
      if ( c[x] == c[y] ) 
	z.push_back(e);
      else {
	int aa = c[x], bb = c[y];
	if ( c[x] == -1 ) aa = K;
	if ( c[y] == -1 ) bb = K;
	if ( !uf.same(aa, bb) ) {
	  uf.unite(aa, bb);
	  z.push_back(e);
	}
      }
      continue;
    }

    if ( !uf.same(c[x], c[y]) ) {
      cnt++;
      ans += w;
      uf.unite(c[x], c[y]);
      used[c[x]] = true;
      used[c[y]] = true;      
    }

    if ( cnt+z.size() >= K-1 ) {
      break;
    }
  }

  for ( int i = 0; i < z.size(); i++ ) {
    cnt++;
    ans += z[i].cost;
    if ( cnt >= K-1 ) break;
  }

  if ( cnt < K-1 ) cout << -1 << endl;
  else cout << ans << endl;
  
  return 0;
}