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;
vector<int> g[2][200010];
int u[2][200010], ccSize[2][200010];
void dfs(int t, int v, int cc) {
  u[t][v] = cc;
  ccSize[t][cc]++;
  for (int i = 0; i < g[t][v].size(); i++) {
    int to = g[t][v][i];
    if (!u[t][to]) {
      dfs(t, to, cc);
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int a, b, t;
    scanf("%d%d%d", &a, &b, &t);
    a--;
    b--;
    g[t][a].push_back(b);
    g[t][b].push_back(a);
  }
  int cc[2] = {};
  for (int i = 0; i < n; i++) {
    if (!u[0][i]) dfs(0, i, ++cc[0]);
    if (!u[1][i]) dfs(1, i, ++cc[1]);
  }
  long long res = 0;
  for (int i = 1; i <= cc[0]; i++) {
    res += 1LL * ccSize[0][i] * (ccSize[0][i] - 1);
  }
  for (int i = 1; i <= cc[1]; i++) {
    res += 1LL * ccSize[1][i] * (ccSize[1][i] - 1);
  }
  for (int i = 0; i < n; i++) {
    res += 1LL * (ccSize[0][u[0][i]] - 1) * (ccSize[1][u[1][i]] - 1);
  }
  printf("%lld", res);
}
