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 + 5;
int f1[N], f2[N];
int Find1(int t) { return f1[t] = (t == f1[t] ? t : Find1(f1[t])); }
int Find2(int t) { return f2[t] = (t == f2[t] ? t : Find2(f2[t])); }
void join1(int x, int y) {
  int fx = Find1(x);
  int fy = Find1(y);
  if (fx != fy) f1[fy] = fx;
}
void join2(int x, int y) {
  int fx = Find2(x);
  int fy = Find2(y);
  if (fx != fy) f2[fy] = fx;
}
int aa[N], bb[N];
long long ans;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) f1[i] = f2[i] = i;
  for (int i = 1; i < n; i++) {
    int x, y, f;
    cin >> x >> y >> f;
    if (f)
      join1(x, y);
    else
      join2(x, y);
  }
  for (int i = 1; i <= n; i++) aa[Find1(i)]++, bb[Find2(i)]++;
  for (int i = 1; i <= n; i++) {
    if (aa[i]) ans += 1ll * aa[i] * (aa[i] - 1);
    if (bb[i]) ans += 1ll * bb[i] * (bb[i] - 1);
    if (aa[f1[i]] && bb[f2[i]]) ans += 1ll * (aa[f1[i]] - 1) * (bb[f2[i]] - 1);
  }
  cout << ans << endl;
  return 0;
}
