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 N = 2e5 + 7;
int fa1[N], fa2[N], cnt1[N], cnt2[N], n;
int find1(int x) {
  if (fa1[x] == x) return x;
  return fa1[x] = find1(fa1[x]);
}
void join1(int x, int y) {
  int tx = find1(x), ty = find1(y);
  if (tx != ty) fa1[tx] = ty;
}
int find2(int x) {
  if (fa2[x] == x) return x;
  return fa2[x] = find2(fa2[x]);
}
void join2(int x, int y) {
  int tx = find2(x), ty = find2(y);
  if (tx != ty) fa2[tx] = ty;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) fa1[i] = fa2[i] = i;
  for (int i = 1; i < n; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    if (z == 0)
      join1(x, y);
    else
      join2(x, y);
  }
  for (int i = 1; i <= n; i++) cnt1[find1(i)]++, cnt2[find2(i)]++;
  long long ans = 0;
  for (int i = 1; i <= n; i++)
    ans += (1ll * cnt1[find1(i)] * cnt2[find2(i)]) - 1;
  cout << ans;
}
