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;
pair<pair<int, int>, int> E[2000005];
int n;
vector<int> adj[2000005];
int R[2000005][2], dR[2000005][2];
bool av[2000005][2];
int findset(int x, int val) {
  if (R[x][val] == x) return x;
  return (R[x][val] = findset(R[x][val], val));
}
void read() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int x, y, z;
    cin >> x >> y >> z;
    E[i] = {{x, y}, z};
  }
  for (int i = 1; i <= n; ++i) R[i][0] = R[i][1] = i, dR[i][0] = dR[i][1] = 1;
}
void solve() {
  for (int i = 1; i < n; ++i) {
    int x, y, z, fx, fy;
    x = E[i].first.first;
    y = E[i].first.second;
    z = E[i].second;
    fx = findset(x, z);
    fy = findset(y, z);
    if (x == y) continue;
    dR[fx][z] += dR[fy][z];
    R[fy][z] = fx;
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    int x = findset(i, 0);
    if (av[x][0] == false) {
      ans += 1ll * dR[x][0] * (dR[x][0] - 1);
      av[x][0] = true;
    }
    int y = findset(i, 1);
    if (av[y][1] == false) {
      ans += 1ll * dR[y][1] * (dR[y][1] - 1);
      av[y][1] = true;
    }
    ans += 1ll * (dR[x][0] - 1) * (dR[y][1] - 1);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  read();
  solve();
  return 0;
}
