You are managing a mobile phone network, and want to offer competitive prices to connect a network.

The network has n nodes.

Your competitor has already offered some connections between some nodes, with some fixed prices. These connections are bidirectional. There are initially m connections the competitor is offering. The i-th connection your competitor is offering will connect nodes fa_i and fb_i and costs fw_i. 

You have a list of k connections that you want to offer. It is guaranteed that this set of connection does not form any cycle. The j-th of these connections will connect nodes ga_j and gb_j. These connections are also bidirectional. The cost of these connections have not been decided yet.

You can set the prices of these connections to any arbitrary integer value. These prices are set independently for each connection. After setting the prices, the customer will choose such n - 1 connections that all nodes are connected in a single network and the total cost of chosen connections is minimum possible. If there are multiple ways to choose such networks, the customer will choose an arbitrary one that also maximizes the number of your connections in it.

You want to set prices in such a way such that all your k connections are chosen by the customer, and the sum of prices of your connections is maximized.

Print the maximum profit you can achieve, or -1 if it is unbounded.

Input

The first line of input will contain three integers n, k and m (1 ≤ n, k, m ≤ 5 ⋅ 10^5, k ≤ n-1), the number of nodes, the number of your connections, and the number of competitor connections, respectively.

The next k lines contain two integers ga_i and gb_i (1 ≤ ga_i, gb_i ≤ n, ga_i not= gb_i), representing one of your connections between nodes ga_i and gb_i. Your set of connections is guaranteed to be acyclic.

The next m lines contain three integers each, fa_i, fb_i and fw_i (1 ≤ fa_i, fb_i ≤ n, fa_i not= fb_i, 1 ≤ fw_i ≤ 10^9), denoting one of your competitor's connections between nodes fa_i and fb_i with cost fw_i. None of these connections connects a node to itself, and no pair of these connections connect the same pair of nodes. In addition, these connections are given by non-decreasing order of cost (that is, fw_{i-1} ≤ fw_i for all valid i).

Note that there may be some connections that appear in both your set and your competitor's set (though no connection will appear twice in one of this sets).

It is guaranteed that the union of all of your connections and your competitor's connections form a connected network.

Output

Print a single integer, the maximum possible profit you can achieve if you set the prices on your connections appropriately. If the profit is unbounded, print -1.

Examples

Input

4 3 6
1 2
3 4
1 3
2 3 3
3 1 4
1 2 4
4 2 8
4 3 8
4 1 10


Output

14


Input

3 2 1
1 2
2 3
1 2 30


Output

-1


Input

4 3 3
1 2
1 3
1 4
4 1 1000000000
4 2 1000000000
4 3 1000000000


Output

3000000000

Note

In the first sample, it's optimal to give connection 1-3 cost 3, connection 1-2 cost 3, and connection 3-4 cost 8. In this case, the cheapest connected network has cost 14, and the customer will choose one that chooses all of your connections.

In the second sample, as long as your first connection costs 30 or less, the customer chooses both your connections no matter what is the cost of the second connection, so you can get unbounded profit in this case.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 10;
int n, m, k, par[maxn], h[maxn], p[maxn], bound[maxn], ga[maxn], gb[maxn],
    fa[maxn], fb[maxn], w[maxn];
bool marked[maxn];
vector<int> g[maxn], newVers;
int root(int v) { return (par[v] == v ? v : par[v] = root(par[v])); }
void merge(int v, int u) {
  v = root(v);
  u = root(u);
  if (v == u) return;
  par[v] = u;
}
void dfs(int v) {
  newVers.push_back(v);
  for (int u : g[v]) {
    if (u != p[v]) {
      h[u] = h[v] + 1;
      p[u] = v;
      dfs(u);
    }
  }
}
int main() {
  cin >> n >> k >> m;
  for (int i = 1; i <= n; i++) par[i] = i;
  for (int i = 0; i < k; i++) {
    scanf("%d%d", &ga[i], &gb[i]);
    merge(ga[i], gb[i]);
  }
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &fa[i], &fb[i], &w[i]);
    if (root(fa[i]) != root(fb[i])) {
      marked[i] = true;
      merge(fa[i], fb[i]);
    }
  }
  for (int i = 1; i <= n; i++) par[i] = i;
  for (int i = 0; i < m; i++) {
    if (marked[i]) merge(fa[i], fb[i]);
  }
  for (int i = 0; i < k; i++) {
    ga[i] = root(ga[i]);
    gb[i] = root(gb[i]);
    g[ga[i]].push_back(gb[i]);
    g[gb[i]].push_back(ga[i]);
  }
  for (int i = 0; i < m; i++) {
    fa[i] = root(fa[i]);
    fb[i] = root(fb[i]);
  }
  for (int i = 1; i <= n; i++) par[i] = i;
  dfs(ga[0]);
  memset(bound, 0x3f, sizeof bound);
  for (int i = 0; i < m; i++) {
    int v = root(fa[i]), u = root(fb[i]);
    while (v != u) {
      if (h[v] < h[u]) swap(v, u);
      bound[v] = w[i];
      par[v] = root(p[v]);
      v = par[v];
    }
  }
  long long ans = 0;
  int inf = 1e9;
  for (int v : newVers)
    if (v != ga[0]) {
      if (bound[v] > inf) {
        cout << -1 << endl;
        return 0;
      }
      ans += 1LL * bound[v];
    }
  cout << ans << endl;
  return 0;
}
