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 sz[200001][2];
int p[200001][2];
long long kq;
int root(int u, int t) {
  if (p[u][t] == u) return u;
  return p[u][t] = root(p[u][t], t);
}
void merges(int u, int v, int t) {
  int ru = root(u, t);
  int rv = root(v, t);
  if (sz[ru][t] < sz[rv][t]) swap(ru, rv);
  p[rv][t] = ru;
  sz[ru][t] += sz[rv][t];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    sz[i][0] = sz[i][1] = 1;
    p[i][0] = p[i][1] = i;
  }
  for (int i = 1; i < n; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    merges(u, v, w);
  }
  for (int i = 1; i <= n; i++) {
    if (i == p[i][0]) kq += sz[i][0] * (sz[i][0] - 1);
    if (i == p[i][1]) kq += sz[i][1] * (sz[i][1] - 1);
  }
  for (int i = 1; i <= n; i++) {
    int r0 = root(i, 0);
    int r1 = root(i, 1);
    kq += (sz[r0][0] - 1) * (sz[r1][1] - 1);
  }
  cout << kq;
  return 0;
}
