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;
const int N = 2e5 + 7;
int par[2][N];
int siz[2][N];
int fnd(int x, int c) {
  if (x == par[c][x]) return x;
  return par[c][x] = fnd(par[c][x], c);
}
void unify(int x, int y, int c) {
  x = fnd(x, c);
  y = fnd(y, c);
  if (siz[c][x] > siz[c][y]) swap(siz[c][x], siz[c][y]);
  par[c][x] = y;
  siz[c][y] += siz[c][x];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n;
  for (int i = 0; i < n; i++) {
    par[0][i] = par[1][i] = i;
    siz[0][i] = siz[1][i] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    --x;
    --y;
    unify(x, y, c);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (par[0][i] == i) ans += siz[0][i] * 1ll * (siz[0][i] - 1);
    if (par[1][i] == i) ans += siz[1][i] * 1ll * (siz[1][i] - 1);
    ans += (siz[0][fnd(i, 0)] - 1) * 1ll * (siz[1][fnd(i, 1)] - 1);
  }
  cout << ans << '\n';
}
