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>
const int N = 200200;
const long long mod = 1e9 + 7;
using namespace std;
int n;
int d[N][2];
long long res;
vector<pair<int, int> > v[N];
void dfs(int x, int p) {
  for (auto y : v[x]) {
    if (y.first == p) {
      continue;
    }
    dfs(y.first, x);
    if (y.second == 1) {
      d[x][1] += d[y.first][1] + 1;
    } else {
      d[x][0] += d[y.first][0] + d[y.first][1] + 1;
    }
  }
  res += d[x][0] + d[x][1];
}
void trace(int x, int p, int a, int b) {
  res += a + b;
  for (auto y : v[x]) {
    if (y.first == p) {
      continue;
    }
    if (y.second == 1) {
      b += d[y.first][1] + 1;
    } else {
      a += d[y.first][0] + d[y.first][1] + 1;
    }
  }
  for (auto y : v[x]) {
    if (y.first == p) {
      continue;
    }
    if (y.second == 1) {
      b -= d[y.first][1] + 1;
    } else {
      a -= d[y.first][0] + d[y.first][1] + 1;
    }
    int na = a, nb = b;
    if (y.second == 1) {
      nb += 1;
      na = 0;
    } else {
      na += 1 + nb;
      nb = 0;
    }
    trace(y.first, x, na, nb);
    if (y.second == 1) {
      b += d[y.first][1] + 1;
    } else {
      a += d[y.first][0] + d[y.first][1] + 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y, w;
    cin >> x >> y >> w;
    v[x].push_back({y, w});
    v[y].push_back({x, w});
  }
  dfs(1, 1);
  trace(1, 1, 0, 0);
  cout << res << "\n";
}
