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 long double PI = acos(-1.0);
const long long N = 2e5 + 4;
vector<pair<long long, long long> > adjlist[N];
long long dp[N][4];
vector<long long> clist[N];
long long ans = 0;
long long contrib[N][4];
void dfs(long long node, long long par) {
  for (auto i : adjlist[node]) {
    if (i.first != par) {
      dfs(i.first, node);
      if (i.second) {
        dp[node][3] += dp[i.first][3] + 1;
        contrib[i.first][3] = dp[i.first][3] + 1;
        dp[node][1] += dp[i.first][1] + dp[i.first][0];
        contrib[i.first][1] = dp[i.first][1] + dp[i.first][0];
      } else {
        dp[node][0] += dp[i.first][0] + 1;
        contrib[i.first][0] = dp[i.first][0] + 1;
        dp[node][2] += dp[i.first][3] + dp[i.first][2];
        contrib[i.first][2] = dp[i.first][3] + dp[i.first][2];
      }
      clist[node].push_back(i.first);
    }
  }
  long long iska = 0;
  iska += 2 * dp[node][3];
  iska += 2 * dp[node][0];
  iska += dp[node][1];
  iska += dp[node][2];
  vector<long long> cur(4, 0);
  for (long long i : clist[node]) {
    iska += 2 * contrib[i][3] * cur[3];
    iska += contrib[i][3] * cur[1];
    iska += contrib[i][3] * cur[0];
    iska += contrib[i][1] * cur[3];
    iska += contrib[i][2] * cur[0];
    iska += 2 * contrib[i][0] * cur[0];
    iska += contrib[i][0] * cur[2];
    iska += contrib[i][0] * cur[3];
    for (long long j = 0; j < 4; ++j) {
      cur[j] += contrib[i][j];
    }
  }
  ans += iska;
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; ++i) {
    long long u, v, w;
    cin >> u >> v >> w;
    adjlist[u].push_back(make_pair(v, w));
    adjlist[v].push_back(make_pair(u, w));
  }
  dfs(1, 0);
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
