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 long N = 200001;
long long n, i, j, k, dem1[N], dem0[N], pos1[N], pos0[N];
long long res;
vector<pair<long long, long long> > adj[N];
vector<long long> sub[N], sub1[N];
bool vis[N], vis1[N];
void dfs(long long u) {
  sub[k].push_back(u);
  vis[u] = true;
  for (auto& x : adj[u]) {
    if (!vis[x.first] && x.second == 0) dfs(x.first);
  }
}
void dfs1(long long u) {
  sub1[j].push_back(u);
  vis1[u] = true;
  for (auto& x : adj[u]) {
    if (!vis1[x.first] && x.second == 1) dfs1(x.first);
  }
}
signed main() {
  cin >> n;
  for (i = 1; i <= n - 1; i++) {
    long long x, y, z;
    cin >> x >> y >> z;
    adj[x].push_back({y, z});
    adj[y].push_back({x, z});
  }
  for (i = 1; i <= n; i++) {
    k++;
    if (!vis[i]) dfs(i);
  }
  for (i = 1; i <= k; i++) {
    long long x = sub[i].size();
    res += x * (x - 1);
  }
  for (i = 1; i <= n; i++) {
    j++;
    if (!vis1[i]) dfs1(i);
  }
  for (i = 1; i <= j; i++) {
    long long x = sub1[i].size();
    res += x * (x - 1);
  }
  for (i = 1; i <= k; i++) {
    for (auto& x : sub[i]) {
      dem0[x] = sub[i].size();
      pos0[x] = i;
    }
  }
  for (i = 1; i <= j; i++) {
    for (auto& x : sub1[i]) {
      dem1[x] = sub1[i].size();
    }
  }
  for (i = 1; i <= n; i++) {
    res += (dem0[i] - 1) * (dem1[i] - 1);
  }
  cout << res;
}
