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;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> VI;
typedef vector<bool> VB;
typedef vector<ll> VL;
const char nl = '\n';
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ld EPS = 1e-9;
const ld PI = acos(-1);
class DSU {
  vector<int> p;
  int c;

 public:
  DSU(int n) : p(n, -1), c(n) {}
  int find(int x) { return (p[x] < 0) ? x : p[x] = find(p[x]); }
  ll size(int x) { return -p[find(x)]; }
  int count() { return c; }
  void join(int x, int y) {
    if ((x = find(x)) == (y = find(y))) return;
    ;
    if (p[x] < p[y]) swap(x, y);
    p[y] += p[x];
    p[x] = y;
    c--;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  DSU edge0(n), edge1(n);
  int x, y, c;
  for (int i = 0; i < n - 1; ++i) {
    cin >> x >> y >> c;
    x--;
    y--;
    if (c) {
      edge1.join(x, y);
    } else {
      edge0.join(x, y);
    }
  }
  ll ans = 0;
  set<int> used0, used1;
  for (int i = 0; i < n; ++i) {
    if (edge0.size(i) > 1 && edge1.size(i) > 1) {
      ans += (edge0.size(i) - 1) * (edge1.size(i) - 1);
    }
    int p = edge0.find(i);
    if (!used0.count(p)) {
      used0.insert(p);
      ans += edge0.size(p) * (edge0.size(p) - 1);
    }
    p = edge1.find(i);
    if (!used1.count(p)) {
      used1.insert(p);
      ans += edge1.size(p) * (edge1.size(p) - 1);
    }
  }
  cout << ans << nl;
  return 0;
}
