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 double pi = acos(-1.0);
int n;
vector<pair<int, int> > p2[200010];
long long ans = 0;
tuple<long long, long long, long long, long long> dfs(int v, int fa) {
  long long ret0 = 0;
  long long ret01 = 0;
  long long ret10 = 0;
  long long ret1 = 0;
  for (auto item : p2[v]) {
    if (item.first == fa) continue;
    auto r = dfs(item.first, v);
    if (item.second == 0) {
      long long add01 = get<1>(r) - 1 + get<3>(r);
      long long add0 = get<0>(r);
      ans += add01 + 2 * add0 + ret0 * add01 + add0 * ret1 + 2 * add0 * ret0 +
             add0 * ret01;
      ret01 += add01;
      ret0 += add0;
    } else {
      long long add10 = get<0>(r) - 1 + get<2>(r);
      long long add1 = get<3>(r);
      ans += add10 + 2 * add1 + ret1 * add10 + add1 * ret0 + 2 * add1 * ret1 +
             add1 * ret10;
      ret10 += add10;
      ret1 += add1;
    }
  }
  ret1++;
  ret0++;
  return {ret0, ret01, ret10, ret1};
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b, w;
    scanf("%d %d %d", &a, &b, &w);
    p2[a].push_back({b, w});
    p2[b].push_back({a, w});
  }
  dfs(1, 0);
  cout << ans << endl;
  return 0;
}
