You are given a tree (an undirected connected acyclic graph) consisting of n vertices and n - 1 edges. A number is written on each edge, each number is either 0 (let's call such edges 0-edges) or 1 (those are 1-edges).

Let's call an ordered pair of vertices (x, y) (x ≠ y) valid if, while traversing the simple path from x to y, we never go through a 0-edge after going through a 1-edge. Your task is to calculate the number of valid pairs in the tree.

Input

The first line contains one integer n (2 ≤ n ≤ 200000) — the number of vertices in the tree.

Then n - 1 lines follow, each denoting an edge of the tree. Each edge is represented by three integers x_i, y_i and c_i (1 ≤ x_i, y_i ≤ n, 0 ≤ c_i ≤ 1, x_i ≠ y_i) — the vertices connected by this edge and the number written on it, respectively.

It is guaranteed that the given edges form a tree.

Output

Print one integer — the number of valid pairs of vertices.

Example

Input


7
2 1 1
3 2 0
4 2 1
5 2 0
6 7 1
7 2 1


Output


34

Note

The picture corresponding to the first example:

<image>
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr int kMaxN = 2e5 + 3;
vector<int> one[kMaxN];
vector<int> zero[kMaxN];
int cur_w;
int w_id[kMaxN];
int w_s[kMaxN];
bitset<kMaxN> used;
void find_w(int u, int p, vector<int>& nodes) {
  nodes.emplace_back(u);
  for (int v : one[u]) {
    if (v != p) {
      find_w(v, u, nodes);
    }
  }
}
void find_b(int u, int p, vector<int>& nodes, vector<int>& other) {
  nodes.emplace_back(u);
  for (int v : zero[u]) {
    if (v != p) {
      find_b(v, u, nodes, other);
    }
  }
  for (int v : one[u]) {
    other.emplace_back(w_id[v]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int u, v, c;
    cin >> u >> v >> c;
    --u, --v;
    if (c == 1) {
      one[u].emplace_back(v);
      one[v].emplace_back(u);
    } else {
      zero[u].emplace_back(v);
      zero[v].emplace_back(u);
    }
  }
  ll total = 0;
  for (int i = 0; i < n; ++i) {
    if (!used[i]) {
      vector<int> nodes;
      find_w(i, -1, nodes);
      for (int v : nodes) {
        used[v] = true;
        w_id[v] = cur_w;
      }
      w_s[cur_w] = nodes.size();
      ++cur_w;
      ll k = nodes.size();
      total += k * (k - 1);
    }
  }
  used.reset();
  for (int i = 0; i < n; ++i) {
    if (!used[i]) {
      vector<int> nodes;
      vector<int> ones;
      find_b(i, -1, nodes, ones);
      for (int v : nodes) {
        used[v] = true;
      }
      sort(ones.begin(), ones.end());
      ones.resize(unique(ones.begin(), ones.end()) - ones.begin());
      ll k = nodes.size();
      total += k * (k - 1);
      for (int w : ones) {
        total += (k - 1) * (w_s[w] - 1);
      }
    }
  }
  cout << total << '\n';
  return 0;
}
