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 oo = 1e9 + 7;
const int modulo = 1e9 + 7;
template <typename T>
inline T sum(const T& a, const T& b) {
  return T((a % modulo + b % modulo) % modulo);
}
template <typename T>
inline T mul(const T& a, const T& b) {
  return T((1LL * (a % (1LL * modulo)) * (b % (1LL * modulo))) %
           (1LL * modulo));
}
string to_string(string a) { return "\"" + a + "\""; }
const int N = 2e5 + 5;
int n, d[2][N], sz[2][N];
int Find(int color, int u) {
  if (d[color][u] == u) return u;
  return d[color][u] = Find(color, d[color][u]);
}
bool Union(int color, int u, int v) {
  u = Find(color, u);
  v = Find(color, v);
  if (u == v) return false;
  sz[color][v] += sz[color][u];
  d[color][u] = v;
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    d[0][i] = d[1][i] = i;
    sz[0][i] = sz[1][i] = 1;
  }
  for (int i = 1; i <= n - 1; ++i) {
    int u, v, c;
    cin >> u >> v >> c;
    Union(c, u, v);
  }
  long long ans = 0;
  for (int u = 1; u <= n; ++u) {
    int t0 = Find(0, u), t1 = Find(1, u);
    if (t0 == u) ans += 1LL * sz[0][u] * (sz[0][u] - 1);
    if (t1 == u) ans += 1LL * sz[1][u] * (sz[1][u] - 1);
    ans += 1LL * (sz[0][t0] - 1) * (sz[1][t1] - 1);
  }
  cout << ans;
  return 0;
}
