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;
using ll = long long;
using pii = pair<int, int>;
struct dsu {
  struct node {
    int p;
    int s;
  };
  vector<node> v;
  dsu(int n) {
    v = vector<node>(n);
    for (int i = 0; i < n; i++) {
      v[i].p = i;
      v[i].s = 1;
    }
  }
  int parent(int s) {
    assert(s < v.size());
    if (v[s].p == s) return s;
    return v[s].p = parent(v[s].p);
  }
  int size(int s) {
    assert(s < v.size());
    return v[parent(s)].s;
  }
  void merge(int a, int b) {
    a = parent(a);
    b = parent(b);
    if (a != b) {
      if (v[a].s < v[b].s) swap(a, b);
      v[b].p = a;
      v[a].s += v[b].s;
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  dsu o(n);
  dsu z(n);
  for (int i = 0; i < n - 1; i++) {
    int x, y, t;
    cin >> x >> y >> t;
    x--;
    y--;
    if (t == 0)
      z.merge(x, y);
    else
      o.merge(x, y);
  }
  ll ans = 0;
  for (int i = 0; i < n; i++) {
    if (o.parent(i) == i) {
      ans += 1LL * o.size(i) * (o.size(i) - 1);
    }
  }
  for (int i = 0; i < n; i++) {
    if (z.parent(i) == i) {
      ans += 1LL * z.size(i) * (z.size(i) - 1);
    }
    ans += 1LL * (o.size(i) - 1) * (z.size(i) - 1);
  }
  cout << ans << '\n';
}
