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;
long long int Lpow(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b % 2) ans = (ans * a);
    a = (a * a);
    b = b / 2;
  }
  return ans;
}
long long int gcd(long long int a, long long int b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
vector<vector<pair<long long int, long long int>>> adj(200005);
vector<long long int> dp0(200005);
vector<long long int> dp1(200005);
vector<long long int> dp0p(200005);
vector<long long int> dp1p(200005);
long long int ans;
void dfs(long long int node, long long int p) {
  for (auto x : adj[node]) {
    if (x.first != p) dfs(x.first, node);
  }
  for (auto x : adj[node]) {
    if (x.first != p) {
      if (x.second == 0) {
        ans += (dp0[node] + 1) * (dp0[x.first] + 1);
        dp0[node] += (dp0[x.first] + 1);
      } else {
        ans += (dp1[node] + 1) * (dp1[x.first] + 1);
        dp1[node] += (dp1[x.first] + 1);
      }
    }
  }
}
void dfs2(long long int node, long long int p, long long int con) {
  if (node == 1) {
    for (auto x : adj[node]) {
      dfs2(x.first, node, x.second);
    }
  } else {
    if (con == 0)
      dp0p[node] = dp0[p] - dp0[node];
    else
      dp1p[node] = dp1[p] - dp1[node];
    dp0[node] = dp0p[node] + dp0[node];
    dp1[node] = dp1p[node] + dp1[node];
    for (auto x : adj[node])
      if (x.first != p) dfs2(x.first, node, x.second);
  }
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int n;
  cin >> n;
  ans = 0;
  long long int a, b, c;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b >> c;
    adj[a].push_back({b, c});
    adj[b].push_back({a, c});
  }
  dfs(1, 1);
  dfs2(1, 1, 0);
  ans = 2 * ans;
  for (int i = 1; i <= n; i++) {
    ans += dp0[i] * dp1[i];
  }
  cout << ans << '\n';
  return 0;
}
