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 {
  vector<int> p, h, s;
  dsu(int n) {
    p.resize(n);
    h.resize(n, 0);
    s.resize(n, 1);
    for (int i = 0; i < n; ++i) {
      p[i] = i;
    }
  }
  int top(int v) {
    if (p[v] == v) return v;
    return p[v] = top(p[v]);
  }
  void unite(int a, int b) {
    a = top(a), b = top(b);
    if (a == b) return;
    if (h[a] < h[b])
      swap(a, b);
    else if (h[a] == h[b])
      h[a]++;
    p[b] = a;
    s[a] += s[b];
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  dsu wg(n), bg(n);
  for (int i = 0; i < n - 1; ++i) {
    int a, b, w;
    cin >> a >> b >> w, --a, --b;
    if (w == 0)
      wg.unite(a, b);
    else
      bg.unite(a, b);
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    int bs = bg.s[bg.top(i)];
    int ws = wg.s[wg.top(i)];
    ans += bs - 1;
    ans += (ws - 1LL) * bs;
  }
  cout << ans << '\n';
}
