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 long long mod = 1e9 + 7;
const int maxn = 2e5 + 5;
vector<pair<int, int> > v[maxn];
long long ans = 0;
long long endp[2][maxn], startdp[2][maxn];
void dfs(int id, int par) {
  for (auto i : v[id]) {
    if (i.first == par) continue;
    dfs(i.first, id);
  }
  for (auto i : v[id]) {
    if (i.first == par) continue;
    if (i.second) {
      endp[1][id] += 1 + endp[1][i.first] + endp[0][i.first];
      startdp[1][id] += 1 + startdp[1][i.first];
    } else {
      endp[0][id] += 1 + endp[0][i.first];
      startdp[0][id] += 1 + startdp[0][i.first] + startdp[1][i.first];
    }
  }
  for (int i = 0; i < 2; ++i) {
    ans += startdp[i][id];
    ans += endp[i][id];
  }
  for (auto i : v[id]) {
    if (i.first == par) continue;
    if (i.second) {
      long long x = 1 + startdp[1][i.first];
      long long y = 1 + endp[1][i.first] + endp[0][i.first];
      ans += y * (startdp[1][id] - x);
    } else {
      long long x = 1 + startdp[0][i.first] + startdp[1][i.first];
      long long y = 1 + endp[0][i.first];
      ans += y * (startdp[0][id] + startdp[1][id] - x);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int x, y, c;
    cin >> x >> y >> c;
    v[x].emplace_back(y, c);
    v[y].emplace_back(x, c);
  }
  dfs(1, -1);
  cout << ans << '\n';
  ;
  return 0;
}
