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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const int INF = 1 << 29;
const int MOD = 1e9 + 7;
const int N = 2e5 + 10;
vector<pair<int, int> > a[N];
int f[2][N], c[2][N];
int find(int f[], int x) {
  if (f[x] != f[f[x]]) f[x] = find(f, f[x]);
  return f[x];
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    --x;
    --y;
    a[x].push_back({y, z});
  }
  long long ret = 0;
  for (int k = 0; k < 2; ++k) {
    for (int i = 0; i < n; ++i) {
      f[k][i] = i;
      c[k][i] = 1;
    }
    for (int i = 0; i < n; ++i) {
      for (auto& it : a[i]) {
        int j = it.first;
        if (it.second != k) continue;
        int ri = find(f[k], i), rj = find(f[k], j);
        if (ri == rj) continue;
        f[k][rj] = ri;
        c[k][ri] += c[k][rj];
      }
    }
    for (int i = 0; i < n; ++i) {
      if (find(f[k], i) == i) {
        ret += (long long)c[k][i] * (c[k][i] - 1);
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    int x = find(f[0], i), y = find(f[1], i);
    ret += (long long)(c[0][x] - 1) * (c[1][y] - 1);
  }
  printf("%lld\n", ret);
  return 0;
}
