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;
void JIZZ(string output = "") {
  cout << output;
  exit(0);
}
const long double PI = 3.14159265358979323846264338327950288;
const long double eps = 1e-10;
const long long mod = 1e9 + 7;
int djs[2][200005], sz[2][200005];
int F(int i, int u) { return djs[i][u] == u ? u : djs[i][u] = F(i, djs[i][u]); }
void U(int i, int u, int v) {
  u = F(i, u);
  v = F(i, v);
  if (u == v) return;
  djs[i][v] = u;
  sz[i][u] += sz[i][v];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    djs[0][i] = djs[1][i] = i;
    sz[0][i] = sz[1][i] = 1;
  }
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    U(w, u, v);
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans += 1ll * (sz[0][F(0, i)] - 1) * (sz[1][F(1, i)] - 1) +
           (sz[0][F(0, i)] - 1) + (sz[1][F(1, i)] - 1);
  }
  cout << ans << endl;
}
