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 INF = 1000 * 1000 * 1000;
const long long LINF = 1LL * INF * INF;
const long double PI = acos(-1.);
const double EPS = 1e-6;
long long mod = INF + 7;
const int maxn = 200010;
int cnt[2][maxn];
vector<int> g[2][maxn];
bool u[2][maxn];
vector<int> comp;
void dfs(int c, int v) {
  comp.push_back(v);
  u[c][v] = 1;
  for (int i = 0; i < (int)g[c][v].size(); ++i) {
    int to = g[c][v][i];
    if (u[c][to]) continue;
    dfs(c, to);
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    g[c][a].push_back(b);
    g[c][b].push_back(a);
  }
  for (int v = 0; v < n; ++v) {
    for (int c = 0; c < 2; ++c) {
      if (!u[c][v]) {
        comp.clear();
        dfs(c, v);
        for (int i = 0; i < (int)comp.size(); ++i)
          cnt[c][comp[i]] = (int)comp.size();
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    ans += cnt[0][i] * (long long)cnt[1][i];
  }
  ans -= n;
  cout << ans;
  return 0;
}
