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;
vector<vector<vector<long long>>> G;
vector<vector<long long>> sz, ccs;
void dfs(const long long &index, const long long &u, const long long &cc) {
  ccs[index][u] = cc;
  ++sz[index][cc];
  for (const long long &v : G[index][u])
    if (ccs[index][v] == -1) dfs(index, v, cc);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long N;
  cin >> N;
  G.resize(2, vector<vector<long long>>(N));
  sz.assign(2, vector<long long>(N));
  ccs.assign(2, vector<long long>(N, -1));
  for (long long i = 1; i < N; ++i) {
    long long u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    G[w][u].emplace_back(v);
    G[w][v].emplace_back(u);
  }
  long long ans = 0, cc1 = 0, cc2 = 0;
  for (long long i = 0; i < N; ++i) {
    if (ccs[0][i] == -1) dfs(0, i, cc1++);
    if (ccs[1][i] == -1) dfs(1, i, cc2++);
    ans += sz[0][ccs[0][i]] * sz[1][ccs[1][i]];
  }
  cout << ans - N << '\n';
}
