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 MX = 2e5;
vector<long long> forest[MX][2];
long long cnt[MX][2];
bool vis[MX][2];
vector<long long> nodes;
void dfs(long long u, long long w) {
  nodes.push_back(u);
  vis[u][w] = true;
  for (long long v : forest[u][w]) {
    if (!vis[v][w]) {
      dfs(v, w);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 0; i < (long long)n - 1; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    forest[u][w].push_back(v);
    forest[v][w].push_back(u);
  }
  long long ans = 0;
  long long full = 0;
  for (int u = 0; u < n; u++) {
    for (long long b = 0; b <= 1; b++) {
      if (!vis[u][b]) {
        nodes.clear();
        dfs(u, b);
        for (long long v : nodes) {
          cnt[v][b] = (long long)nodes.size() - 1;
        }
      }
    }
    full += cnt[u][0] + cnt[u][1];
    ans += cnt[u][0] * cnt[u][1];
  }
  ans += full;
  cout << ans << "\n";
  return 0;
}
