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 T>
inline void re(T &x) {
  char c;
  bool nega = 0;
  while ((!isdigit(c = getchar())) && (c != '-'))
    ;
  if (c == '-') {
    nega = 1;
    c = getchar();
  }
  x = c - 48;
  while (isdigit(c = getchar())) x = x * 10 + c - 48;
  if (nega) x = -x;
}
template <typename T>
inline void writep(T x) {
  if (x > 9) writep(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  writep(x);
}
template <typename T>
inline void writeln(T x) {
  write(x);
  putchar('\n');
}
vector<int> sizes[2], par[2];
int find_par(int x, int c) {
  if (par[c][x] == x) return x;
  return par[c][x] = find_par(par[c][x], c);
}
void Union_(int x, int y, int c) {
  x = find_par(x, c);
  y = find_par(y, c);
  if (sizes[c][x] < sizes[c][y]) swap(x, y);
  par[c][y] = x;
  sizes[c][x] += sizes[c][y];
}
int main() {
  int n;
  re(n);
  for (int i = 0; i < 2; ++i) {
    sizes[i].resize(n, 1);
    for (int j = 0; j < n; ++j) par[i].push_back(j);
  }
  for (int i = 0; i + 1 < n; ++i) {
    int x, y, c;
    re(x), re(y), re(c);
    --x, --y;
    Union_(x, y, c);
  }
  auto res = 0ll;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 2; ++j)
      if (par[j][i] == i) {
        res += 1ll * (sizes[j][i] - 1) * sizes[j][i];
      }
    res +=
        1ll * (sizes[0][find_par(i, 0)] - 1) * (sizes[1][find_par(i, 1)] - 1);
  }
  printf("%lld", res);
  return 0;
}
