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;
using cd = complex<double>;
const long long INF = 1e18;
const long long mod = 998244353;
const double Pi = acos(-1);
void Fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int n;
int dsu[200005][2], sz[200005][2];
int Root(int node, int id) {
  return dsu[node][id] == node ? node : dsu[node][id] = Root(dsu[node][id], id);
}
signed main() {
  Fastio();
  cin >> n;
  for (int i = 1; i <= n; i++) {
    dsu[i][0] = i;
    dsu[i][1] = i;
    sz[i][0] = 1;
    sz[i][1] = 1;
  }
  for (int i = 1; i <= n - 1; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    sz[Root(v, c)][c] += sz[Root(u, c)][c];
    dsu[Root(u, c)][c] = Root(v, c);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += 1ll * sz[Root(i, 0)][0] * sz[Root(i, 1)][1];
    --ans;
  }
  cout << ans;
}
