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 = 3e5;
int n, u, v, tp;
int par[2][N], sz[2][N];
long long ans;
int find_set(int v) {
  if (par[tp][v] == v) return v;
  return par[tp][v] = find_set(par[tp][v]);
}
void union_sets(int a, int b) {
  a = find_set(a);
  b = find_set(b);
  if (sz[tp][a] < sz[tp][b]) swap(a, b);
  par[tp][b] = a;
  sz[tp][a] += sz[tp][b];
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    par[0][i] = par[1][i] = i;
    sz[0][i] = sz[1][i] = 1;
  }
  for (int i = 1; i < n; ++i) {
    cin >> u >> v >> tp;
    --u;
    --v;
    union_sets(u, v);
  }
  for (int i = 0; i < n; ++i) {
    if (par[0][i] == i) ans += (long long)sz[0][i] * (sz[0][i] - 1);
    if (par[1][i] == i) ans += (long long)sz[1][i] * (sz[1][i] - 1);
    tp = 0;
    int v = find_set(i);
    tp = 1;
    int u = find_set(i);
    ans += (long long)(sz[0][v] - 1) * (sz[1][u] - 1);
  }
  cout << ans;
}
