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 int N = 2e5;
int sz[N + 2], tot, cnt[N + 2], flag[N + 2], ok[3], mark[N + 2];
vector<pair<int, int> > adj[N + 2];
long long ans = 0;
void reckon(int node, int par) {
  sz[node] = 1;
  tot++;
  for (auto x : adj[node]) {
    if (x.first == par || mark[x.first]) continue;
    reckon(x.first, node);
    sz[node] += sz[x.first];
  }
}
int find_center(int node, int par) {
  for (auto x : adj[node]) {
    if (x.first == par || mark[x.first]) continue;
    if (sz[x.first] > tot / 2) return find_center(x.first, node);
  }
  return node;
}
void cal(int bss, int node, int par, int f) {
  cnt[bss]++;
  for (auto x : adj[node]) {
    if (x.first == par || mark[x.first]) continue;
    if (!x.second && f) continue;
    cal(bss, x.first, node, x.second);
  }
}
void agn(int bss, int node, int par, int l, int o) {
  ans++;
  if (o) {
    ans += ok[1] * 1LL;
    if (flag[bss]) ans -= cnt[bss];
  } else {
    ans += (ok[0] + ok[1] - cnt[bss]) * 1LL;
  }
  for (auto x : adj[node]) {
    if (x.first == par || mark[x.first]) continue;
    if (!l && x.second) continue;
    agn(bss, x.first, node, x.second, o | x.second);
  }
}
void centroid(int node) {
  tot = 0;
  reckon(node, -1);
  int center = find_center(node, -1);
  mark[center] = 1;
  ok[0] = 0;
  ok[1] = 0;
  for (auto x : adj[center]) {
    if (mark[x.first]) continue;
    flag[x.first] = x.second;
    cnt[x.first] = 0;
    cal(x.first, x.first, -1, x.second);
    ok[flag[x.first]] += cnt[x.first];
  }
  ans += (ok[0] + ok[1]) * 1LL;
  for (auto x : adj[center]) {
    if (mark[x.first]) continue;
    agn(x.first, x.first, -1, x.second, x.second);
  }
  for (auto x : adj[center]) {
    if (mark[x.first]) continue;
    centroid(x.first);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    adj[u].push_back({v, w});
    adj[v].push_back({u, w});
  }
  centroid(1);
  cout << ans << endl;
  return 0;
}
