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;
int n;
long long ans;
struct Set {
  int pa[200005], size[200005];
  void pre() {
    for (int i = 1; i <= n; i++) pa[i] = i, size[i] = 1;
  }
  int find(int x) { return pa[x] == x ? x : pa[x] = find(pa[x]); }
  void merge(int x, int y) {
    x = find(x), y = find(y);
    if (x != y) pa[x] = y, size[y] += size[x];
  }
  void calc() {
    for (int i = 1; i <= n; i++)
      if (find(i) == i) ans += 1ll * size[i] * (size[i] - 1);
  }
} A, B;
int num[200005];
int main() {
  scanf("%d", &n);
  A.pre(), B.pre();
  for (int i = 1; i < n; i++) {
    int u, v, l;
    scanf("%d%d%d", &u, &v, &l);
    if (l == 0)
      A.merge(u, v);
    else
      B.merge(u, v);
  }
  A.calc();
  B.calc();
  for (int i = 1; i <= n; i++)
    ans += 1ll * (B.size[B.find(i)] - 1) * (A.size[A.find(i)] - 1);
  printf("%I64d\n", ans);
  return 0;
}
