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;
unsigned long long ans, n, one[200030], zero[200030];
vector<pair<unsigned long long, unsigned long long> > v[200030];
void dfs1(int node, int par) {
  for (auto pp : v[node]) {
    if (pp.first == par) continue;
    dfs1(pp.first, node);
    if (pp.second == 0)
      zero[node] += zero[pp.first] + one[pp.first] + 1;
    else
      one[node] += one[pp.first] + 1;
  }
}
void dfs2(int node, int par, int val, bool checked) {
  ans += val + one[node] + zero[node];
  for (auto pp : v[node]) {
    if (pp.first == par) continue;
    if (!checked) {
      if (pp.second)
        dfs2(pp.first, node, one[node] - (one[pp.first] + 1) + val + 1, 0);
      if (!pp.second)
        dfs2(pp.first, node,
             val + zero[node] + one[node] - (zero[pp.first] + one[pp.first]),
             1);
    } else {
      if (pp.second) dfs2(pp.first, node, one[node] - one[pp.first], 0);
      if (!pp.second)
        dfs2(pp.first, node,
             val + zero[node] + one[node] - (zero[pp.first] + one[pp.first]),
             1);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    v[x].push_back({y, c});
    v[y].push_back({x, c});
  }
  dfs1(1, -1);
  dfs2(1, -1, 0, 0);
  cout << ans << endl;
  return 0;
}
