#include <bits/stdc++.h>
const int maxn = 212345;
using namespace std;
int fa[2][maxn], siz[2][maxn];
int root(int x, int z) {
  if (fa[z][x] == x)
    return x;
  else
    return fa[z][x] = root(fa[z][x], z);
}
void zhan(int x, int y, int z) {
  x = root(x, z);
  y = root(y, z);
  if (siz[z][x] < siz[z][y]) swap(x, y);
  fa[z][y] = x;
  siz[z][x] += siz[z][y];
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i <= n; i++) {
    fa[0][i] = fa[1][i] = i;
    siz[0][i] = siz[1][i] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    zhan(u, v, w);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (fa[0][i] == i) ans += siz[0][i] * 1ll * (siz[0][i] - 1);
    if (fa[1][i] == i) ans += siz[1][i] * 1ll * (siz[1][i] - 1);
    ans += (siz[0][root(i, 0)] - 1) * 1ll * (siz[1][root(i, 1)] - 1);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
const int maxn = 212345;
using namespace std;
int fa[2][maxn], siz[2][maxn];
int root(int x, int z) {
  if (fa[z][x] == x)
    return x;
  else
    return fa[z][x] = root(fa[z][x], z);
}
void zhan(int x, int y, int z) {
  x = root(x, z);
  y = root(y, z);
  if (siz[z][x] < siz[z][y]) swap(x, y);
  fa[z][y] = x;
  siz[z][x] += siz[z][y];
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i <= n; i++) {
    fa[0][i] = fa[1][i] = i;
    siz[0][i] = siz[1][i] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    zhan(u, v, w);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (fa[0][i] == i) ans += siz[0][i] * 1ll * (siz[0][i] - 1);
    if (fa[1][i] == i) ans += siz[1][i] * 1ll * (siz[1][i] - 1);
    ans += (siz[0][root(i, 0)] - 1) * 1ll * (siz[1][root(i, 1)] - 1);
  }
  cout << ans << endl;
  return 0;
}

```