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 n, u, v, c, Cnt0 = 0, Cnt1 = 0, degree[2][200001], O[200001];
long long res = 0, X, Y;
vector<int> LT[2][200001], ds[2][200001];
bool F[2][200001];
void DFS(int c, int u, int Cnt) {
  if (c == 1) O[u] = Cnt;
  F[c][u] = 1;
  LT[c][Cnt].push_back(u);
  for (int i = 0; i < ds[c][u].size(); i++) {
    int v = ds[c][u][i];
    if (F[c][v] == 0) {
      degree[c][u]--;
      degree[c][v]--;
      DFS(c, v, Cnt);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d%d%d", &u, &v, &c);
    ds[c][u].push_back(v);
    ds[c][v].push_back(u);
    degree[c][u]++;
    degree[c][v]++;
  }
  for (int i = 1; i <= n; i++) {
    if (degree[0][i] > 0) DFS(0, i, ++Cnt0);
    if (degree[1][i] > 0) DFS(1, i, ++Cnt1);
  }
  for (int i = 1; i <= Cnt0; i++)
    res += (long long)(LT[0][i].size() - 1) * LT[0][i].size();
  for (int i = 1; i <= Cnt1; i++)
    res += (long long)(LT[1][i].size() - 1) * LT[1][i].size();
  for (int i = 1; i <= Cnt0; i++) {
    for (int j = 0; j < LT[0][i].size(); j++) {
      int k = LT[0][i][j];
      if (O[k] != 0) {
        X = LT[0][i].size() - 1;
        Y = LT[1][O[k]].size() - 1;
        res += X * Y;
      }
    }
  }
  printf("%lld", res);
}
