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;
vector<pair<int, int>> adj[200005];
pair<long long, long long> tmp, dp[200005], w[200005];
int vis[200005], T = 1;
long long rlt;
pair<long long, long long> DFS(int s) {
  vis[s] = T;
  pair<long long, long long> ret;
  ret.first = 0, ret.second = 0;
  for (int i = adj[s].size() - 1; i >= 0; i--) {
    int v = adj[s][i].first;
    if (vis[v] == T) continue;
    tmp = DFS(v);
    if (adj[s][i].second == 0) {
      ret.first += (tmp.first + tmp.second + 1);
    } else {
      ret.second += tmp.second + 1;
    }
  }
  w[s] = ret;
  return ret;
}
void DFS1(int s) {
  vis[s] = T;
  for (int i = adj[s].size() - 1; i >= 0; i--) {
    int v = adj[s][i].first;
    if (vis[v] == T) continue;
    if (adj[s][i].second == 1) {
      tmp.first = w[v].first;
      tmp.second = dp[s].second;
    } else {
      tmp.first = (dp[s].first + dp[s].second - w[v].second);
      tmp.second = w[v].second;
    }
    dp[v] = tmp;
    DFS1(v);
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v, val;
    cin >> u >> v >> val;
    adj[u - 1].push_back(make_pair(v - 1, val));
    adj[v - 1].push_back(make_pair(u - 1, val));
  }
  DFS(0);
  dp[0] = w[0];
  T++;
  DFS1(0);
  for (int i = 0; i < n; i++) {
    rlt += (dp[i].first + dp[i].second);
  }
  cout << rlt << endl;
  return 0;
}
