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;
template <class A, class B>
ostream& operator<<(ostream& out, const pair<A, B>& a) {
  return out << "(" << a.first << ", " << a.second << ")";
}
template <class A>
ostream& operator<<(ostream& out, const vector<A>& v) {
  out << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) out << ", ";
    out << v[i];
  }
  return out << "]";
}
const int N = 200005;
int n;
int p[2][N];
int cnt[2][N];
int Find(int x, int c) {
  if (p[c][x] == x) return x;
  return p[c][x] = Find(p[c][x], c);
}
void Union(int x, int y, int c) {
  int px = Find(x, c);
  int py = Find(y, c);
  if (cnt[c][px] < cnt[c][py]) swap(px, py);
  p[c][py] = px;
  cnt[c][px] += cnt[c][py];
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cnt[1][i] = cnt[0][i] = 1;
    p[1][i] = p[0][i] = i;
  }
  for (int i = 1; i < n; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    Union(x, y, c);
  }
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    long long Z = cnt[0][Find(i, 0)];
    long long O = cnt[1][Find(i, 1)];
    if (p[0][i] == i) res += (Z * (Z - 1));
    if (p[1][i] == i) res += (O * (O - 1));
    res += (Z - 1) * (O - 1);
  }
  cout << res << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q = 1;
  for (int kase = 1; kase <= q; kase++) {
    solve();
  }
  return 0;
}
