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;
struct dsu {
  int f[200005], size[200005];
  inline void reset(int n) {
    for (int i = 1; i <= n; i++) f[i] = i, size[i] = 1;
  }
  inline int getf(int x) { return f[x] != x ? f[x] = getf(f[x]) : x; }
  inline void merge(int x, int y) {
    x = getf(x), y = getf(y);
    size[x] += size[y];
    f[y] = x;
  }
  inline int query(int x) {
    x = getf(x);
    return size[x];
  }
} dsu0, dsu1;
int main() {
  int n;
  cin >> n;
  dsu0.reset(n);
  dsu1.reset(n);
  for (int i = 1; i < n; i++) {
    int x, y, w;
    cin >> x >> y >> w;
    if (w == 0)
      dsu0.merge(x, y);
    else
      dsu1.merge(x, y);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int s, t;
    s = dsu0.query(i);
    t = dsu1.query(i);
    if (i == dsu0.getf(i)) ans += 1LL * s * (s - 1);
    if (i == dsu1.getf(i)) ans += 1LL * t * (t - 1);
    ans += 1LL * (s - 1) * (t - 1);
  }
  cout << ans;
  return 0;
}
