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;
long long n, ans = 0;
long long father[2][200005], size[2][200005];
struct node {
  long long u, v, w;
} a[200005];
long long C(long long x) { return (x * x - x) / 2; }
long long find(long long t, long long x) {
  if (x == father[t][x]) return x;
  return father[t][x] = find(t, father[t][x]);
}
void Merge(long long x, long long y, long long flag) {
  long long fx = find(flag, x), fy = find(flag, y);
  father[flag][fx] = fy;
  size[flag][fy] += size[flag][fx];
  size[flag][fx] = 0;
}
signed main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) father[0][i] = i;
  for (long long i = 1; i < n; i++) cin >> a[i].u >> a[i].v >> a[i].w;
  for (long long i = 1; i <= n; i++) father[0][i] = i, size[0][i] = 1;
  for (long long i = 1; i < n; i++) {
    if (a[i].w == 0) Merge(a[i].u, a[i].v, 0);
  }
  for (long long i = 1; i <= n; i++) father[1][i] = i, size[1][i] = 1;
  for (long long i = 1; i < n; i++) {
    if (a[i].w == 1) Merge(a[i].u, a[i].v, 1);
  }
  for (long long i = 1; i <= n; i++) {
    long long fx = find(0, i);
    long long fy = find(1, i);
    ans += size[0][fx] * size[1][fy];
  }
  cout << ans - n << endl;
  return 0;
}
