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;
const int oo = 1e9 + 5;
int du[] = {-1, 0, 0, 1};
int dv[] = {0, -1, 1, 0};
const long long mod = 1e9 + 7;
long long sqr(long long x) { return x * x; }
struct data {
  int u, v, w;
  data(int U = 0, int V = 0, int W = 0) { u = U, v = V, w = W; }
} edge[N + 5];
int n, dad[N + 5][2], s[N + 5][2];
long long res;
int getdad(int u, int k) {
  if (dad[u][k] == u) return u;
  return dad[u][k] = getdad(dad[u][k], k);
}
void join(int u, int v, int k) {
  int p = getdad(u, k), q = getdad(v, k);
  if (p == q) return;
  if (s[p][k] > s[q][k]) swap(p, q);
  dad[p][k] = q;
  s[q][k] += s[p][k];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    edge[i] = data(u, v, w);
  }
  for (int i = 1; i <= n; i++) {
    dad[i][0] = i;
    s[i][0] = 1;
    dad[i][1] = i;
    s[i][1] = 1;
  }
  for (int i = 1; i < n; i++) {
    if (edge[i].w == 0)
      join(edge[i].u, edge[i].v, 0);
    else
      join(edge[i].u, edge[i].v, 1);
  }
  for (int i = 1; i <= n; i++) {
    if (getdad(i, 0) == i) res += 1ll * s[i][0] * (s[i][0] - 1);
    if (getdad(i, 1) == i) res += 1ll * s[i][1] * (s[i][1] - 1);
    int p = getdad(i, 0), q = getdad(i, 1);
    res += 1ll * (s[p][0] - 1) * (s[q][1] - 1);
  }
  cout << res;
  return 0;
}
