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 p[2][200001];
long long int sz[2][200001];
int get_parent(int x, int c) {
  if (p[c][x] == x) return x;
  return p[c][x] = get_parent(p[c][x], c);
}
void merge(int x, int y, int c) {
  int px = get_parent(x, c);
  int py = get_parent(y, c);
  if (px == py) return;
  if (sz[c][py] < sz[c][px]) swap(px, py);
  p[c][px] = py;
  sz[c][py] += sz[c][px];
}
int main() {
  int n, x, y, c;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    p[0][i] = p[1][i] = i;
    sz[0][i] = sz[1][i] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d%d", &x, &y, &c);
    x--;
    y--;
    merge(x, y, c);
  }
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    int p0 = get_parent(i, 0);
    int p1 = get_parent(i, 1);
    long long int s0 = sz[0][p0];
    long long int s1 = sz[1][p1];
    ans += (s0 - 1) * (s1 - 1);
    if (i == p0) ans += s0 * (s0 - 1);
    if (i == p1) ans += s1 * (s1 - 1);
  }
  cout << ans << "\n";
}
