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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 610031;
int n;
vector<pair<int, int> > edges[3];
int w[N];
int SZ0[N];
int SZ1[N];
long long SZ01[N];
int sz[N];
long long ans;
int get(int x) {
  if (x == w[x]) return x;
  return w[x] = get(w[x]);
}
int root0[N];
void merge(int a, int b) {
  a = get(a);
  b = get(b);
  w[a] = b;
  sz[b] += sz[a];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b, clr;
    cin >> a >> b >> clr;
    edges[clr].push_back(make_pair(a, b));
  }
  for (int i = 1; i <= n; i++) {
    w[i] = i;
    sz[i] = 1;
  }
  for (int i = 0; i < edges[1].size(); i++) {
    int a = edges[1][i].first;
    int b = edges[1][i].second;
    merge(a, b);
  }
  for (int i = 1; i <= n; i++) {
    SZ1[i] = sz[get(i)];
  }
  for (int i = 1; i <= n; i++) {
    w[i] = i;
    sz[i] = 1;
  }
  for (int i = 0; i < edges[0].size(); i++) {
    int a = edges[0][i].first;
    int b = edges[0][i].second;
    merge(a, b);
  }
  for (int i = 1; i <= n; i++) {
    int a = get(i);
    SZ01[a] += SZ1[i];
  }
  for (int i = 1; i <= n; i++) {
    ans += SZ01[get(i)];
  }
  cout << ans - n << endl;
  return 0;
}
