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 = 2e5 + 10;
bool vis[N];
struct Edge {
  int v, w;
};
vector<Edge> G[N];
vector<int> tmp, block0, block1;
int cnt0[N], cnt1[N];
void dfs(int u, int f, int w) {
  vis[u] = 1;
  tmp.push_back(u);
  for (auto &e : G[u]) {
    if (e.v == f || e.w != w) continue;
    dfs(e.v, u, w);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    G[u].push_back(Edge{v, w});
    G[v].push_back(Edge{u, w});
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      tmp.clear();
      dfs(i, 0, 0);
      block0.push_back(tmp.size());
      for (auto &j : tmp) cnt0[j] = tmp.size();
    }
  }
  memset(vis, 0, sizeof vis);
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      tmp.clear();
      dfs(i, 0, 1);
      block1.push_back(tmp.size());
      for (auto &j : tmp) cnt1[j] = tmp.size();
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += 1LL * (cnt0[i] - 1) * (cnt1[i] - 1);
  }
  for (auto &i : block0) ans += 1LL * i * (i - 1);
  for (auto &i : block1) ans += 1LL * i * (i - 1);
  cout << ans << endl;
  return 0;
}
