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;
const long long N = 2e5 + 5;
vector<long long> g0[N], g1[N], sz0(N), sz1(N), temp;
bool vis[N];
void dfs0(long long v, long long p) {
  vis[v] = 1;
  temp.push_back(v);
  for (auto &it : g0[v]) {
    if (p != it) dfs0(it, v);
  }
  return;
}
void dfs1(long long v, long long p) {
  vis[v] = 1;
  temp.push_back(v);
  for (auto &it : g1[v]) {
    if (p != it) dfs1(it, v);
  }
  return;
}
int main() {
  int n;
  cin >> n;
  long long ans = 0;
  for (int i = 0; i < n - 1; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    if (!w)
      g0[u].push_back(v), g0[v].push_back(u);
    else
      g1[u].push_back(v), g1[v].push_back(u);
  }
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      temp.clear();
      dfs0(i, 0);
      long long l = temp.size();
      for (auto it : temp) sz0[it] = l;
      ans += (l * (l - 1));
    }
  }
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      temp.clear();
      dfs1(i, 0);
      long long l = temp.size();
      for (auto it : temp) sz1[it] = l;
      ans += (l * (l - 1));
    }
  }
  for (int i = 1; i <= n; i++) ans += (sz0[i] - 1) * (sz1[i] - 1);
  cout << ans;
}
