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 MAXN = 200002;
int n;
int p[2][MAXN], sz[2][MAXN];
int root(int nodo, int p[]) {
  if (p[nodo] == 0) return nodo;
  return p[nodo] = root(p[nodo], p);
}
void join(int a, int b, int p[], int sz[]) {
  int ra = root(a, p), rb = root(b, p);
  p[ra] = rb;
  sz[rb] += sz[ra];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  fill(sz[0] + 1, sz[0] + n + 1, 1);
  fill(sz[1] + 1, sz[1] + n + 1, 1);
  for (int i = 0; i < n - 1; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    join(a, b, p[c], sz[c]);
  }
  long long int ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans += sz[0][root(i, p[0])] - 1;
    ans += sz[1][root(i, p[1])] - 1;
    ans += (long long int)(sz[0][root(i, p[0])] - 1) *
           (long long int)(sz[1][root(i, p[1])] - 1);
  }
  cout << ans << "\n";
  return 0;
}
