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 mod = 1e9 + 7;
const int inf = 1e9;
int n;
vector<pair<int, int>> e[200005];
int a[200005][2];
long long ans;
void dfs(int u, int f) {
  for (auto pp : e[u]) {
    if (pp.first == f) continue;
    int v = pp.first;
    dfs(v, u);
    if (pp.second == 0) {
      a[u][0] += a[v][0];
    } else {
      a[u][1] += a[v][1];
      a[u][1] += a[v][0];
    }
  }
  a[u][0] += 1;
}
void get(int u, int f, int s0, int s1) {
  s0 += a[u][0];
  s1 += a[u][1];
  ans += s0 + s1;
  for (auto pp : e[u]) {
    if (pp.first == f) continue;
    int v = pp.first;
    if (pp.second == 0) {
      get(v, u, s0 - a[v][0], 0);
    } else {
      get(v, u, 0, s0 + s1 - a[v][0] - a[v][1]);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int u, v, x;
    scanf("%d%d%d", &u, &v, &x);
    u--;
    v--;
    e[u].push_back({v, x});
    e[v].push_back({u, x});
  }
  int root = 0;
  dfs(root, root);
  get(root, root, 0, 0);
  cout << ans - n << endl;
  return 0;
}
