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 n;
vector<vector<pair<int, int>>> gr;
vector<int> comp;
vector<long long> val{0};
vector<pair<long long, long long>> c;
int cur = 1;
int dfs(int u, int type) {
  comp[u] = cur;
  int res = 1;
  for (auto i : gr[u]) {
    if (!comp[i.first] && i.second == type) {
      res += dfs(i.first, type);
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int _ = 1;
  while (_--) {
    cin >> n;
    gr.resize(n);
    comp.resize(n);
    c.resize(n);
    for (int i = 0; i < n - 1; i++) {
      int a, b, c;
      cin >> a >> b >> c;
      gr[--a].push_back({--b, c});
      gr[b].push_back({a, c});
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      if (!comp[i]) {
        val.push_back(dfs(i, 0) - 1);
        ans += *val.rbegin() * (*(val.rbegin()) + 1);
        cur++;
      }
    }
    for (int i = 0; i < n; i++) {
      c[i].first = val[comp[i]];
    }
    comp.assign(n, 0);
    cur = 1;
    val = vector<long long>{0};
    for (int i = 0; i < n; i++) {
      if (!comp[i]) {
        val.push_back(dfs(i, 1) - 1);
        ans += *val.rbegin() * (*(val.rbegin()) + 1);
        cur++;
      }
    }
    for (int i = 0; i < n; i++) {
      c[i].second = val[comp[i]];
    }
    for (int i = 0; i < n; i++) {
      ans += c[i].first * c[i].second;
    }
    cout << ans << endl;
  }
}
