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;
int par[3][200000 + 10];
long long sz[3][200000 + 10];
int n;
int find(int n, int c) {
  if (n == par[c][n])
    return n;
  else
    return par[c][n] = find(par[c][n], c);
}
void Union(int a, int b, int c) {
  int x = find(a, c);
  int y = find(b, c);
  if (x != y) {
    if (x > y) swap(x, y);
    par[c][y] = x;
    sz[c][x] += sz[c][y];
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    par[0][i] = par[1][i] = i;
    sz[0][i] = sz[1][i] = 1;
  }
  for (int i = 1; i < n; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    Union(a, b, c);
  }
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    if (par[0][i] == i) res += sz[0][i] * (sz[0][i] - 1);
    if (par[1][i] == i) res += sz[1][i] * (sz[1][i] - 1);
    res += (sz[0][find(i, 0)] - 1) * (sz[1][find(i, 1)] - 1);
  }
  cout << res << endl;
}
