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 = 2e5;
int n, f[N + 5][2];
std::vector<std::pair<int, int>> E[N + 5];
inline void addEdge(int u, int v, int w) {
  E[u].push_back(std::make_pair(v, w));
}
void dfs1(int u, int p) {
  for (const auto &t : E[u]) {
    int v = t.first, w = t.second;
    if (v != p) {
      dfs1(v, u);
      f[u][w]++;
      f[u][w] += (w == 1 ? f[v][1] + f[v][0] : f[v][0]);
    }
  }
}
void dfs2(int u, int p) {
  for (const auto &t : E[u]) {
    int v = t.first, w = t.second;
    if (v != p) {
      f[v][w] +=
          (w == 1 ? f[u][1] - f[v][0] - f[v][1] + f[u][0] : f[u][0] - f[v][0]);
      dfs2(v, u);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int u, v, w, i = 1; i < n; i++) {
    scanf("%d%d%d", &u, &v, &w);
    addEdge(u, v, w);
    addEdge(v, u, w);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += f[i][0] + f[i][1];
  }
  printf("%lld\n", ans);
  return 0;
}
