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 long long Mod = 7 + 1e9;
const long long INF32 = 5 + 2e9;
const long long INF64 = 1 + 5e3;
const long long MAX = 5 + 3e5;
int N, boss[MAX], sz[MAX];
vector<pair<int, bool> > G[MAX];
int DFS1(int u, int pre, int BOSS) {
  boss[u] = BOSS;
  int sum = 0;
  for (auto v : G[u])
    if (v.first != pre && v.second == 1) sum += DFS1(v.first, u, BOSS) + 1;
  return sum;
}
int cnt;
bool vis[MAX];
int DFS2(int u, int pre) {
  vis[u] = 1;
  cnt += sz[boss[u]];
  int sum = 1;
  for (auto v : G[u])
    if (v.first != pre && v.second == 0) sum += DFS2(v.first, u);
  return sum;
}
int main() {
  cin >> N;
  for (int i = 0; i < N - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    G[u].push_back({v, w});
    G[v].push_back({u, w});
  }
  long long ans = 0;
  for (int i = 1; i <= N; i++)
    if (boss[i] == 0) sz[i] = DFS1(i, i, i);
  for (int i = 1; i <= N; i++)
    if (!vis[i]) {
      cnt = 0;
      long long temp = DFS2(i, i);
      ans += temp * cnt;
      ans += temp * (temp - 1);
    }
  cout << ans << endl;
  return 0;
}
