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;
long long p[2][200005], sz[2][200005], i = 0, n, x, y, z, ans = 0;
long long get(long long a, long long c) {
  if (p[c][a] == a) return a;
  return get(p[c][a], c);
}
void merge(long long a, long long b, long long c) {
  p[c][a] = get(a, c);
  a = p[c][a];
  p[c][b] = get(b, c);
  b = p[c][b];
  if (sz[c][a] < sz[c][b]) swap(a, b);
  sz[c][a] += sz[c][b];
  p[c][b] = a;
}
int main(void) {
  for (cin >> n; i < n; i++) {
    sz[0][i] = sz[1][i] = 1;
    p[0][i] = p[1][i] = i;
  }
  for (i = 0; i < n - 1; i++) {
    cin >> x >> y >> z;
    x--, y--;
    merge(x, y, z);
  }
  for (i = 0; i < n; i++) {
    if (p[0][i] == i) {
      ans += sz[0][i] * (sz[0][i] - 1);
    }
    if (p[1][i] == i) {
      ans += sz[1][i] * (sz[1][i] - 1);
    }
    ans += (sz[0][get(i, 0)] - 1) * (sz[1][get(i, 1)] - 1);
  }
  cout << ans;
  return 0;
}
