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 + 5;
int n;
long long sz[N][2];
int fa[N][2];
int find(int u, int x) {
  return u == fa[u][x] ? u : fa[u][x] = find(fa[u][x], x);
}
inline void Union(int u, int v, int x) {
  int f1 = find(u, x), f2 = find(v, x);
  if (f1 != f2) {
    fa[f2][x] = f1;
    sz[f1][x] += sz[f2][x];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    sz[i][0] = 1, sz[i][1] = 1;
    fa[i][0] = i, fa[i][1] = i;
  }
  for (int i = 1; i < n; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    Union(u, v, w);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (fa[i][0] == i) ans += sz[i][0] * (sz[i][0] - 1);
    if (fa[i][1] == i) ans += sz[i][1] * (sz[i][1] - 1);
    int f1 = find(i, 1), f0 = find(i, 0);
    ans += (sz[f0][0] - 1) * (sz[f1][1] - 1) * 1LL;
  }
  cout << ans << endl;
}
