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;
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  int x, y, c;
  vector<int> neigh0[n + 1];
  vector<int> neigh1[n + 1];
  long long zeros[n + 1];
  long long ones[n + 1];
  bool done[n + 1];
  for (int i = 1; i < n; i++) {
    cin >> x >> y >> c;
    if (c == 0) {
      neigh0[x].push_back(y);
      neigh0[y].push_back(x);
    } else {
      neigh1[x].push_back(y);
      neigh1[y].push_back(x);
    }
  }
  for (int i = 0; i <= n; i++) {
    done[i] = false;
  }
  for (int i = 1; i <= n; i++) {
    if (!done[i]) {
      int curr = i;
      long long connected = 0;
      queue<int> bfs;
      queue<int> been;
      been.push(curr);
      for (int j = 0; j < neigh0[curr].size(); j++) {
        bfs.push(neigh0[curr][j]);
        done[neigh0[curr][j]] = 1;
        been.push(neigh0[curr][j]);
        connected++;
      }
      done[curr] = 1;
      while (!bfs.empty()) {
        curr = bfs.front();
        bfs.pop();
        for (int j = 0; j < neigh0[curr].size(); j++) {
          if (!done[neigh0[curr][j]]) {
            bfs.push(neigh0[curr][j]);
            done[neigh0[curr][j]] = 1;
            been.push(neigh0[curr][j]);
            connected++;
          }
        }
      }
      while (!been.empty()) {
        int temp = been.front();
        been.pop();
        zeros[temp] = connected;
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    done[i] = false;
  }
  for (int i = 1; i <= n; i++) {
    if (!done[i]) {
      int curr = i;
      int connected = 0;
      queue<int> bfs;
      queue<int> been;
      been.push(curr);
      for (int j = 0; j < neigh1[curr].size(); j++) {
        bfs.push(neigh1[curr][j]);
        done[neigh1[curr][j]] = 1;
        been.push(neigh1[curr][j]);
        connected++;
      }
      done[curr] = 1;
      while (!bfs.empty()) {
        curr = bfs.front();
        bfs.pop();
        for (int j = 0; j < neigh1[curr].size(); j++) {
          if (!done[neigh1[curr][j]]) {
            bfs.push(neigh1[curr][j]);
            done[neigh1[curr][j]] = 1;
            been.push(neigh1[curr][j]);
            connected++;
          }
        }
      }
      while (!been.empty()) {
        int temp = been.front();
        been.pop();
        ones[temp] = connected;
      }
    }
  }
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    res += (zeros[i] + 1) * (ones[i] + 1) - 1;
  }
  cout << res;
}
