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 int N = 200004;
int parent[2][N];
int size_[2][N];
int find(int x, int c) {
  if (parent[c][x] == x) {
    return x;
  }
  return parent[c][x] = find(parent[c][x], c);
}
void merge(int x, int y, int c) {
  x = find(x, c);
  y = find(y, c);
  if (size_[c][x] < size_[c][y]) {
    swap(x, y);
  }
  parent[c][y] = x;
  size_[c][x] += size_[c][y];
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    parent[0][i] = parent[1][i] = i;
    size_[0][i] = size_[1][i] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    x--, y--;
    merge(x, y, z);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (parent[0][i] == i) {
      ans += (size_[0][i] * 1ll * (size_[0][i] - 1));
    }
    if (parent[1][i] == i) {
      ans += (size_[1][i] * 1ll * (size_[1][i] - 1));
    }
    ans += (size_[0][find(i, 0)] - 1) * 1ll * (size_[1][find(i, 1)] - 1);
  }
  cout << ans << endl;
  return 0;
}
