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;
int n;
struct state {
  int t, c;
};
vector<state> E[200005];
int f[200005][2][2];
long long ans = 0;
void dfs(int fa, int u) {
  int ts[2][2];
  memset(ts, 0, sizeof(ts));
  for (int i = 0; i < E[u].size(); i++) {
    if (E[u][i].t != fa) {
      dfs(u, E[u][i].t);
      if (E[u][i].c == 0) {
        f[u][0][0] += f[E[u][i].t][0][0] + 1;
        f[u][1][0] += (f[E[u][i].t][1][0] + f[E[u][i].t][1][1]);
        f[E[u][i].t][0][0]++;
        f[E[u][i].t][1][0] += f[E[u][i].t][1][1];
        f[E[u][i].t][0][1] = f[E[u][i].t][1][1] = 0;
      } else {
        f[u][1][1] += f[E[u][i].t][1][1] + 1;
        f[u][0][1] += (f[E[u][i].t][0][1] + f[E[u][i].t][0][0]);
        f[E[u][i].t][1][1]++;
        f[E[u][i].t][0][1] += f[E[u][i].t][0][0];
        f[E[u][i].t][0][0] = f[E[u][i].t][1][0] = 0;
      }
      ts[0][0] += f[E[u][i].t][0][0];
      ts[0][1] += f[E[u][i].t][0][1];
      ts[1][0] += f[E[u][i].t][1][0];
      ts[1][1] += f[E[u][i].t][1][1];
    }
  }
  ans += 2 * f[u][0][0] + 2 * f[u][1][1] + f[u][0][1] + f[u][1][0];
  for (int i = 0; i < E[u].size(); i++) {
    if (E[u][i].t == fa) continue;
    ans += (1LL * f[E[u][i].t][0][0] *
            (ts[0][0] - f[E[u][i].t][0][0] + ts[1][0] - f[E[u][i].t][1][0] +
             ts[1][1] - f[E[u][i].t][1][1]));
    ans += (1LL * f[E[u][i].t][0][1] * (ts[1][1] - f[E[u][i].t][1][1]));
    ans += (1LL * f[E[u][i].t][1][1] * (ts[1][1] - f[E[u][i].t][1][1]));
  }
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v, c;
    scanf("%d%d%d", &u, &v, &c);
    E[u].push_back((state){v, c});
    E[v].push_back((state){u, c});
  }
  dfs(0, 1);
  cout << ans;
  return 0;
}
