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 i64 = long long;
void dfs(int u, const vector<vector<int>> &g, vector<int> &vis,
         vector<int> &cur) {
  vis[u] = 1;
  cur.push_back(u);
  for (int v : g[u]) {
    if (!vis[v]) {
      dfs(v, g, vis, cur);
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<vector<int>> g[2];
  g[0].resize(n);
  g[1].resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int x, y, c;
    cin >> x >> y >> c;
    g[c][x - 1].push_back(y - 1);
    g[c][y - 1].push_back(x - 1);
  }
  vector<int> cnt[2];
  cnt[0].resize(n);
  cnt[1].resize(n);
  i64 res = 0;
  for (int c = 0; c < 2; ++c) {
    vector<int> vis(n, 0);
    for (int u = 0; u < n; ++u) {
      if (!vis[u]) {
        vector<int> cur;
        dfs(u, g[c], vis, cur);
        int sz = cur.size();
        res += (i64)sz * (sz - 1);
        for (int v : cur) {
          cnt[c][v] = sz - 1;
        }
      }
    }
  }
  for (int u = 0; u < n; ++u) {
    res += (i64)cnt[0][u] * cnt[1][u];
  }
  cout << res << "\n";
  return 0;
}
