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 pii = pair<int, int>;
using vi = vector<int>;
using ll = long long;
const int N = 2e5;
vector<pair<int, bool>> adj[N];
ll ans = 0;
int c0[N], c1[N], c01[N], c10[N];
void dfs(int u = 0, int p = -1) {
  for (auto [v, b] : adj[u])
    if (v != p) {
      dfs(v, u);
      if (b) {
        ans += (ll)(c0[u] + c01[u] + 2 * c1[u]) * (c1[v] + 1);
        ans += (ll)c1[u] * (c0[v] + c01[v]);
        c01[u] += c0[v] + c01[v];
        c1[u] += c1[v] + 1;
      } else {
        ans += (ll)(c1[u] + c10[u] + 2 * c0[u]) * (c0[v] + 1);
        ans += (ll)c0[u] * (c1[v] + c10[v]);
        c10[u] += c1[v] + c10[v];
        c0[u] += c0[v] + 1;
      }
    }
  ans += c10[u] + c01[u] + 2 * c0[u] + 2 * c1[u];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int _ = int(0); _ < int(n - 1); _++) {
    int u, v;
    bool b;
    cin >> u >> v >> b;
    u--, v--;
    adj[u].emplace_back(v, b);
    adj[v].emplace_back(u, b);
  }
  dfs();
  cout << ans << '\n';
  return 0;
}
