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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 2e5 + 5, M = 2 * N + 5;
int n;
vector<pair<int, int> > edges[2];
long long sz[N][2];
int parent[N][2];
void init() {
  for (int i = 0; i < n; ++i)
    sz[i][0] = sz[i][1] = 1, parent[i][0] = parent[i][1] = i;
}
int find(int u, bool fen) {
  return parent[u][fen] == u ? u : parent[u][fen] = find(parent[u][fen], fen);
}
void merge(int u, int v, bool fen) {
  u = find(u, fen);
  v = find(v, fen);
  if (u == v) return;
  if (sz[u][fen] > sz[v][fen]) swap(u, v);
  parent[u][fen] = parent[v][fen];
  sz[v][fen] += sz[u][fen];
  return;
}
int main() {
  scanf("%d", &n);
  init();
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    merge(--u, --v, w);
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    if (parent[i][0] == i) ans += (sz[i][0] * (sz[i][0] - 1LL));
    if (parent[i][1] == i) ans += (sz[i][1] * (sz[i][1] - 1LL));
    int u = find(i, 0);
    int v = find(i, 1);
    ans += ((sz[u][0] - 1LL) * (sz[v][1] - 1LL));
  }
  printf("%lld\n", ans);
  return 0;
}
