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 INF = 1e18;
const int N = 3e+5 + 5;
long long int cnt[2][N], vis[2][N], visup[2][N];
long long int n, m, k, x, u, v, ans, ct;
vector<pair<long long int, long long int>> adj[N];
string s;
queue<pair<long long int, long long int>> q;
void dfs(long long int i, int key) {
  if (!vis[key][i]) {
    vis[key][i] = 1;
    ct += 1;
    for (auto it : adj[i]) {
      if (it.second == key) dfs(it.first, key);
    }
  }
}
void dfsup(long long int i, int key) {
  if (!visup[key][i]) {
    visup[key][i] = 1;
    cnt[key][i] = ct;
    for (auto it : adj[i]) {
      if (it.second == key) dfsup(it.first, key);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    cin >> u >> v >> x;
    adj[u].push_back({v, x});
    adj[v].push_back({u, x});
  }
  for (int i = 1; i <= n; i++) {
    ct = 0;
    if (!vis[0][i]) {
      dfs(i, 0);
      dfsup(i, 0);
    }
    ans += ct * (ct - 1);
    ct = 0;
    if (!vis[1][i]) {
      dfs(i, 1);
      dfsup(i, 1);
    }
    ans += ct * (ct - 1);
  }
  for (int i = 1; i <= n; i++) {
    ans += (cnt[0][i] - 1) * (cnt[1][i] - 1);
  }
  cout << ans << endl;
}
