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>
int get() {
  int x = 0;
  char s = getchar();
  while (s < '0' || s > '9') s = getchar();
  while (s >= '0' && s <= '9')
    x = (x << 3) + (x << 1) + (s ^ 48), s = getchar();
  return x;
}
const int M = 2e5 + 5;
int n, f[M][2], a[M], b[M];
long long ans;
int find(const int x, const bool y) {
  if (f[x][y] ^ x) f[x][y] = find(f[x][y], y);
  return f[x][y];
}
int main() {
  n = get();
  for (int i = 1; i <= n; ++i) f[i][0] = f[i][1] = i;
  for (int i = 1, u, v, c, fu, fv; i < n; ++i) {
    u = get(), v = get(), c = get();
    fu = find(u, c), fv = find(v, c);
    if (fu ^ fv) f[fu][c] = fv;
  }
  for (int i = 1; i <= n; ++i) ++a[find(i, 0)], ++b[find(i, 1)];
  for (int i = 1; i <= n; ++i) ans += 1ll * a[find(i, 0)] * b[find(i, 1)] - 1;
  printf("%lld", ans);
  return 0;
}
