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 a0[200005], a1[200005], number0[200005], number1[200005];
long long Find(int node, int flag) {
  if (flag == 0) {
    if (node != a0[node]) {
      a0[node] = Find(a0[node], 0);
    }
    return a0[node];
  }
  if (flag == 1) {
    if (node != a1[node]) {
      a1[node] = Find(a1[node], 1);
    }
    return a1[node];
  }
}
void Union(int x, int y, int flag) {
  if (flag == 0) {
    x = Find(x, 0);
    y = Find(y, 0);
    if (x != y) {
      a0[x] = y;
      number0[y] += number0[x];
    }
  }
  if (flag == 1) {
    x = Find(x, 1);
    y = Find(y, 1);
    if (x != y) {
      a1[x] = y;
      number1[y] += number1[x];
    }
  }
}
int main() {
  long long n, a, b, t, sum = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    a0[i] = a1[i] = i;
    number0[i] = number1[i] = 1;
  }
  for (int i = 1; i < n; i++) {
    cin >> a >> b >> t;
    Union(a, b, t);
  }
  for (int i = 1; i <= n; i++) {
    long long num0 = number0[Find(i, 0)] - 1LL,
              num1 = number1[Find(i, 1)] - 1LL;
    sum += (long long)(num0) + (long long)(num1);
    sum += (long long)(num0) * (long long)(num1);
  }
  cout << sum;
  return 0;
}
