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;
const int N = 2e5 + 5;
int n, par[N][2], sz[N][2];
long long ans;
int get_par(int u, int x) {
  return ~par[u][x] ? par[u][x] = get_par(par[u][x], x) : u;
}
void merge(int u, int v, int x) {
  u = get_par(u, x);
  v = get_par(v, x);
  if (u == v) return;
  sz[par[v][x] = u][x] += sz[v][x];
}
int main() {
  cin >> n;
  for (int u = 0; u < n; u++) {
    sz[u][0] = sz[u][1] = 1;
    par[u][0] = par[u][1] = -1;
  }
  for (int i = 1; i < n; i++) {
    int u, v, x;
    cin >> u >> v >> x;
    merge(--u, --v, x);
    merge(v, u, x);
  }
  for (int u = 0; u < n; u++)
    ans += 1LL * sz[get_par(u, 0)][0] * sz[get_par(u, 1)][1];
  cout << ans - n;
}
