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;
struct disjoint_set {
  long long n;
  vector<long long> prt;
  disjoint_set(long long n) : n(n), prt(n, -1) {}
  bool is_root(long long x) { return prt[x] < 0; }
  long long size(long long x) { return -prt[find(x)]; }
  bool equiv(long long x, long long y) { return find(x) == find(y); }
  long long find(long long x) {
    if (is_root(x))
      return x;
    else
      return prt[x] = find(prt[x]);
  }
  bool unite(long long x, long long y) {
    x = find(x);
    y = find(y);
    if (x == y) return false;
    if (size(x) < size(y)) swap(x, y);
    prt[x] += prt[y];
    prt[y] = x;
    return true;
  }
};
int main() {
  long long n;
  cin >> n;
  disjoint_set djs(n), ejs(n);
  vector<pair<long long, long long>> ddg, edg;
  map<pair<long long, long long>, long long> mp;
  for (long long i = 0; i < n - 1; i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    a--, b--;
    if (c == 0) {
      ddg.emplace_back(a, b);
    } else {
      edg.emplace_back(a, b);
    }
  }
  long long ret = 0;
  for (auto e : ddg) {
    long long a, b;
    tie(a, b) = e;
    djs.unite(a, b);
  }
  for (auto e : edg) {
    long long a, b;
    tie(a, b) = e;
    ejs.unite(a, b);
  }
  for (long long i = 0; i < n; i++) {
    if (djs.is_root(i)) {
      long long sz = djs.size(i);
      ret += sz * (sz - 1);
    }
    if (ejs.is_root(i)) {
      long long sz = ejs.size(i);
      ret += sz * (sz - 1);
    }
  }
  for (long long i = 0; i < n; i++) {
    mp[{djs.find(i), ejs.find(i)}]++;
  }
  for (auto it = mp.begin(); it != mp.end(); it++) {
    long long x, y;
    tie(x, y) = it->first;
    long long w = it->second;
    ret += (djs.size(x) - w) * (ejs.size(y) - w);
  }
  cout << ret << endl;
  return 0;
}
