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 par[200002][2];
int sz[200002][2];
int n, m;
int finds(int x, int c) {
  if (par[x][c] == x)
    return x;
  else
    return par[x][c] = finds(par[x][c], c);
}
void joins(int x, int y, int c) {
  int paru = finds(x, c);
  int parv = finds(y, c);
  if (paru != parv) {
    par[paru][c] = par[parv][c];
    sz[parv][c] += sz[paru][c];
  }
}
signed main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 1; j++) {
      par[i][j] = i;
      sz[i][j] = 1;
    }
  }
  m = n - 1;
  while (m--) {
    int x, y, c;
    cin >> x >> y >> c;
    joins(x, y, c);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (i == par[i][0]) ans += 1LL * sz[i][0] * (sz[i][0] - 1);
    if (i == par[i][1]) ans += 1LL * sz[i][1] * (sz[i][1] - 1);
    ans += 1LL * (sz[finds(i, 0)][0] - 1) * (sz[finds(i, 1)][1] - 1);
  }
  cout << ans;
}
