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;
int dsu[2][(200 * 1000 + 5)], sz[2][(200 * 1000 + 5)],
    colors[2][(200 * 1000 + 5)];
int find(int c, int i) {
  return dsu[c][i] == i ? dsu[c][i] : dsu[c][i] = find(c, dsu[c][i]);
}
void unite(int c, int i, int j) {
  i = find(c, i), j = find(c, j);
  if (i == j) return;
  if (sz[c][i] < sz[c][j]) {
    sz[c][j] += sz[c][i];
    dsu[c][i] = j;
  } else {
    sz[c][i] += sz[c][j];
    dsu[c][j] = i;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> edges[2];
  for (int i = 0; i < n - 1; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    edges[c].emplace_back(u, v);
    colors[c][u] = colors[c][v] = 1;
  }
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < (200 * 1000 + 5); j++) {
      dsu[i][j] = j;
      sz[i][j] = 1;
    }
  for (int i = 0; i < 2; i++) {
    for (auto& e : edges[i]) {
      int u = e.first, v = e.second;
      unite(i, u, v);
    }
  }
  long long ans = 0;
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < (200 * 1000 + 5); j++) {
      if (dsu[i][j] == j) {
        ans += ((long long)sz[i][j]) * ((long long)sz[i][j] - 1);
      }
    }
  }
  for (int i = 0; i < (200 * 1000 + 5); i++) {
    if (colors[0][i] & colors[1][i]) {
      int x = find(0, i);
      int y = find(1, i);
      ans += ((long long)sz[0][x] - 1) * ((long long)sz[1][y] - 1);
    }
  }
  cout << ans << endl;
}
