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;
vector<pair<int, int> > g[200100];
bool used[200100][2];
int pr[200100][2];
int cost[200100][2];
int dfs(int v, int type) {
  pr[v][type] = v;
  used[v][type] = 1;
  int ret = 1;
  for (int i = 0; i < g[v].size(); i++) {
    pair<int, int> to = g[v][i];
    if (to.second == type && !used[to.first][to.second]) {
      ret += dfs(to.first, to.second);
      pr[to.first][to.second] = v;
    }
  }
  return ret;
}
int get(int v, int type) {
  while (pr[v][type] != v) {
    pr[v][type] = pr[pr[v][type]][type];
    v = pr[v][type];
  }
  return cost[v][type];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  long long ans = 0;
  for (int i = 1; i <= n - 1; i++) {
    int f, s, c;
    cin >> f >> s >> c;
    g[f].push_back(make_pair(s, c));
    g[s].push_back(make_pair(f, c));
  }
  for (int i = 1; i <= n; i++) {
    long long f, s;
    if (!used[i][1]) {
      s = dfs(i, 1);
      cost[i][1] = s;
    } else
      s = get(i, 1);
    if (!used[i][0]) {
      f = dfs(i, 0);
      cost[i][0] = f;
    } else
      f = get(i, 0);
    ans += (f - 1);
    ans += (s - 1);
    ans += (f - 1) * (s - 1);
  }
  cout << ans << endl;
  return 0;
}
