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 warning(disable : 4996)
using namespace std;
const int N = 200001;
int* f;
int** F;
int** cnt;
inline int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
int main() {
  int n;
  scanf("%d", &n);
  long long ans = 0;
  F = new int*[2];
  F[0] = new int[n + 1];
  F[1] = new int[n + 1];
  cnt = new int*[2];
  cnt[0] = new int[n + 1];
  cnt[1] = new int[n + 1];
  for (int i = 1; i <= n; i++) {
    F[0][i] = F[1][i] = i;
    cnt[0][i] = 0;
    cnt[1][i] = 0;
  }
  for (int i = 1; i < n; i++) {
    int x, y, c;
    scanf("%d %d %d", &x, &y, &c);
    f = F[c];
    if (find(x) != find(y)) f[f[x]] = f[y];
  }
  f = F[0];
  for (int i = 1; i <= n; i++) cnt[0][find(i)]++;
  for (int i = 1; i <= n; i++)
    if (f[i] == i) ans += (long long)cnt[0][i] * (long long)(cnt[0][i] - 1ll);
  f = F[1];
  for (int i = 1; i <= n; i++) cnt[1][find(i)]++;
  for (int i = 1; i <= n; i++)
    if (f[i] == i) ans += (long long)cnt[1][i] * (long long)(cnt[1][i] - 1ll);
  for (int i = 1; i <= n; i++)
    ans +=
        (long long)(cnt[0][F[0][i]] - 1ll) * (long long)(cnt[1][F[1][i]] - 1ll);
  delete F[0];
  delete F[1];
  delete[] F;
  delete cnt[0];
  delete cnt[1];
  delete[] cnt;
  printf("%lld", ans);
  return 0;
}
