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;
long long int power(long long int a, long long int b, long long int m) {
  if (b == 0) return (1);
  long long int sol = power(a, b / 2, m);
  sol = (sol * sol) % m;
  if (b % 2 == 1) sol = (sol * a) % m;
  return (sol);
}
long long int gcd(long long int a, long long int b) {
  if (a % b == 0)
    return b;
  else {
    return gcd(b, a % b);
  }
}
vector<int> g0[200005], g1[200005];
bool used[200005];
int cnt0[200005], cnt1[200005];
vector<int> comp;
void dfs0(int u) {
  used[u] = true;
  comp.push_back(u);
  for (int v : g0[u]) {
    if (!used[v]) dfs0(v);
  }
}
void dfs1(int u) {
  used[u] = true;
  comp.push_back(u);
  for (int v : g1[u]) {
    if (!used[v]) dfs1(v);
  }
}
void solve() {
  int n, u, v, w;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v >> w;
    u--, v--;
    if (w == 0) {
      g0[u].push_back(v);
      g0[v].push_back(u);
    } else {
      g1[u].push_back(v);
      g1[v].push_back(u);
    }
  }
  long long int ans = 0;
  fill(used, used + n, false);
  for (int i = 0; i < n; i++) {
    if (!used[i]) {
      comp.clear();
      dfs0(i);
      ans = ans + 1LL * comp.size() * (comp.size() - 1);
      for (int x : comp) cnt0[x] = comp.size();
    }
  }
  fill(used, used + n, false);
  for (int i = 0; i < n; i++) {
    if (!used[i]) {
      comp.clear();
      dfs1(i);
      ans = ans + 1LL * comp.size() * (comp.size() - 1);
      for (int x : comp) cnt1[x] = comp.size();
    }
  }
  for (int i = 0; i < n; i++) {
    ans = ans + 1LL * (cnt0[i] - 1) * (cnt1[i] - 1);
  }
  cout << ans;
}
int main() {
  int t = 1;
  while (t--) solve();
  return (0);
}
