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, bool>> v[200001];
int jos[200001][2], sus[200001][2];
long long rasp;
void dfs1(int r, int t) {
  int i, rv;
  for (i = 0; i < v[r].size(); i++)
    if (v[r][i].first != t) {
      rv = v[r][i].first;
      dfs1(rv, r);
      if (v[r][i].second == 0)
        jos[r][0] = jos[r][0] + jos[rv][0] + 1;
      else
        jos[r][1] = jos[r][1] + jos[rv][1] + 1;
    }
}
void dfs2(int r, int t) {
  int i, rv;
  for (i = 0; i < v[r].size(); i++)
    if (v[r][i].first != t) {
      rv = v[r][i].first;
      sus[rv][v[r][i].second] = sus[r][v[r][i].second] +
                                jos[r][v[r][i].second] -
                                jos[rv][v[r][i].second];
      rasp = rasp + 1ll * (jos[rv][v[r][i].second] + 1) *
                        (jos[r][v[r][i].second] - jos[rv][v[r][i].second] - 1);
      dfs2(rv, r);
    }
  rasp = rasp + 1ll * sus[r][0] * jos[r][1] + 1ll * sus[r][1] * jos[r][0] +
         1ll * jos[r][0] * jos[r][1];
  rasp = rasp + 2ll * jos[r][0] + 2ll * jos[r][1];
}
int main() {
  int i, n, x, y;
  bool z;
  cin >> n;
  for (i = 1; i < n; i++) {
    cin >> x >> y >> z;
    v[x].push_back({y, z});
    v[y].push_back({x, z});
  }
  dfs1(1, 0);
  dfs2(1, 0);
  cout << rasp;
  return 0;
}
