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, m, par[2][200005];
vector<int> gr[200005];
int root(int v, int flag) {
  return par[flag][v] < 0 ? v : (par[flag][v] = root(par[flag][v], flag));
}
void union_set(int u, int v, int flag) {
  if ((u = root(u, flag)) == (v = root(v, flag))) return;
  if (par[flag][u] > par[flag][v]) swap(u, v);
  par[flag][u] += par[flag][v];
  par[flag][v] = u;
}
vector<pair<int, pair<int, int> > > edges;
set<int> myset;
long long ans = 0;
void dfs(int now, int from) {
  int x1 = root(now, 0);
  int x2 = root(now, 1);
  long long sz1 = -par[0][x1], sz2 = -par[1][x2];
  sz1--;
  sz2--;
  ans += sz1 * sz2;
  for (auto to : gr[now]) {
    if (to != from) dfs(to, now);
  }
}
int main() {
  int i, j;
  memset(par, -1, sizeof(par));
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    int u, v, c;
    scanf("%d", &u);
    scanf("%d", &v);
    scanf("%d", &c);
    edges.push_back({c, {u, v}});
    gr[u].push_back(v);
    gr[v].push_back(u);
  }
  for (auto x : edges) {
    union_set(x.second.first, x.second.second, x.first);
  }
  for (i = 1; i <= n; i++) {
    myset.insert(root(i, 0));
  }
  for (auto x : myset) {
    long long sz = -par[0][x];
    ans += (sz * (sz - 1));
  }
  myset.clear();
  for (i = 1; i <= n; i++) myset.insert(root(i, 1));
  for (auto x : myset) {
    long long sz = -par[1][x];
    ans += (sz * (sz - 1));
  }
  dfs(1, -1);
  cout << ans << endl;
  return 0;
}
