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 Size1[1000010], Size2[1000010], f1[1000010], f2[1000010];
int find1(int x) { return f1[x] ? f1[x] = find1(f1[x]) : x; }
int find2(int x) { return f2[x] ? f2[x] = find2(f2[x]) : x; }
void union1(int x, int y) {
  if ((x = find1(x)) != (y = find1(y))) f1[x] = y, Size1[y] += Size1[x];
}
void union2(int x, int y) {
  if ((x = find2(x)) != (y = find2(y))) f2[x] = y, Size2[y] += Size2[x];
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) Size1[i] = Size2[i] = 1;
  for (int i = 1, x, y, z; i < n; i++) {
    scanf("%d%d%d", &x, &y, &z);
    if (z)
      union2(x, y);
    else
      union1(x, y);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++)
    ans += (long long)Size1[find1(i)] * Size2[find2(i)] - 1;
  printf("%lld\n", ans);
  return 0;
}
