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 + 10;
int n;
long long par[N][2];
long long res = 0;
int root(int u, int type) {
  return par[u][type] < 0 ? u : par[u][type] = root(par[u][type], type);
}
bool join(int u, int v, int type) {
  u = root(u, type);
  v = root(v, type);
  if (u == v) return false;
  if (par[u][type] > par[v][type]) swap(u, v);
  par[u][type] += par[v][type];
  par[v][type] = u;
  return true;
}
int main() {
  scanf("%d", &n);
  memset(par, -1, sizeof(par));
  for (int i = 1; i < n; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    join(x, y, z);
  }
  for (int i = 1; i <= n; i++) {
    long long d0 = -par[root(i, 0)][0], d1 = -par[root(i, 1)][1];
    res += d0 * d1 - 1;
  }
  cout << res;
}
