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 N = 200010;
int n;
vector<int> g0[N];
vector<int> g1[N];
bool used[N];
vector<int> comp;
int sz0[N];
int sz1[N];
void dfs0(int idx) {
  used[idx] = true;
  comp.push_back(idx);
  for (auto adj : g0[idx]) {
    if (!used[adj]) {
      dfs0(adj);
    }
  }
}
void dfs1(int idx) {
  used[idx] = true;
  comp.push_back(idx);
  for (auto adj : g1[idx]) {
    if (!used[adj]) {
      dfs1(adj);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int x, y, c;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y >> c;
    --x;
    --y;
    if (c == 0) {
      g0[x].push_back(y);
      g0[y].push_back(x);
    } else {
      g1[x].push_back(y);
      g1[y].push_back(x);
    }
  }
  long long int ans = 0;
  memset(used, false, sizeof(used));
  for (int i = 0; i < n; i++) {
    comp.clear();
    if (!used[i]) {
      dfs0(i);
      ans += 1LL * comp.size() * (comp.size() - 1);
      for (auto x : comp) {
        sz0[x] = comp.size();
      }
    }
  }
  memset(used, false, sizeof(used));
  for (int i = 0; i < n; i++) {
    comp.clear();
    if (!used[i]) {
      dfs1(i);
      ans += 1LL * comp.size() * (comp.size() - 1);
      for (auto x : comp) {
        sz1[x] = comp.size();
      }
    }
  }
  for (int i = 0; i < n; i++) {
    ans += 1LL * (sz0[i] - 1) * (sz1[i] - 1);
  }
  cout << ans;
  return 0;
}
