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;
vector<pair<int, int> > v[200010];
int edg[200010], c_0[200010], c_1[200010], c_01[200010], c_10[200010];
long long ans;
void dfs(int now, int pv) {
  for (auto x : v[now])
    if (x.first != pv) {
      edg[x.first] = x.second;
      dfs(x.first, now);
    }
}
void count(int now, int pv) {
  for (int i = 0; i < v[now].size(); i++)
    if (v[now][i].first != pv) {
      pair<int, int> x = v[now][i];
      count(x.first, now);
      int t_0 = 0, t_1 = 0, t_01 = 0, t_10 = 0;
      if (edg[x.first]) {
        t_1 = c_1[x.first] + 1;
        t_01 = c_0[x.first] + c_01[x.first];
      } else {
        t_0 = c_0[x.first] + 1;
        t_10 = c_1[x.first] + c_10[x.first];
      }
      ans += 1LL * t_0 * (c_0[now] + c_10[now] + c_1[now]);
      ans += 1LL * c_1[now] * (t_01 + t_1);
      ans += 1LL * c_0[now] * (t_0 + t_10);
      ans += 1LL * t_1 * (c_0[now] + c_01[now] + c_1[now]);
      if (v[x.first].size() == 1 && x.first != 1) {
        if (edg[x.first])
          c_1[now]++;
        else
          c_0[now]++;
      } else {
        if (edg[x.first]) {
          c_01[now] += c_0[x.first] + c_01[x.first];
          c_1[now] += c_1[x.first] + 1;
        } else {
          c_0[now] += c_0[x.first] + 1;
          c_10[now] += c_1[x.first] + c_10[x.first];
        }
      }
    }
  ans += 2LL * (c_0[now] + c_1[now]) + c_01[now] + c_10[now];
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    v[a].push_back({b, c});
    v[b].push_back({a, c});
  }
  dfs(1, -1);
  count(1, -1);
  cout << ans << endl;
}
