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;
long long int n, a, b, type;
vector<vector<pair<long long int, long long int>>> g;
long long int compsize;
vector<bool> visited;
vector<pair<long long int, long long int>> edgesize;
vector<long long int> vertexInComp;
void dfs(int s, int type) {
  visited[s] = true;
  compsize++;
  vertexInComp.push_back(s);
  for (auto it = g[s].begin(); it != g[s].end(); it++) {
    if (!visited[it->first] && (it->second == type)) {
      dfs(it->first, type);
    }
  }
}
int main() {
  cin >> n;
  g.resize(n);
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b >> type;
    a--;
    b--;
    g[a].push_back(make_pair(b, type));
    g[b].push_back(make_pair(a, type));
  }
  visited.resize(n, false);
  edgesize.resize(n);
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    if (!visited[i]) {
      compsize = 0;
      vertexInComp.clear();
      dfs(i, 1);
      for (int j = 0; j < vertexInComp.size(); j++) {
        edgesize[vertexInComp[j]].second = compsize;
      }
      ans += compsize * (compsize - 1);
    }
  }
  visited.clear();
  visited.resize(n, false);
  for (int i = 0; i < n; i++) {
    if (!visited[i]) {
      compsize = 0;
      vertexInComp.clear();
      dfs(i, 0);
      for (int j = 0; j < vertexInComp.size(); j++) {
        edgesize[vertexInComp[j]].first = compsize;
      }
      ans += compsize * (compsize - 1);
    }
  }
  for (int i = 0; i < n; i++) {
    ans += (edgesize[i].first - 1) * (edgesize[i].second - 1);
  }
  cout << ans;
  return 0;
}
