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;
struct Tree {
  vector<vector<int> > be, ge;
  vector<int> bc, gc;
  int n;
  Tree(int n_) : be(n_), ge(n_), bc(n_, 0), gc(n_, 0) { n = n_; }
  void add(int i, int j, int col) {
    if (col == 0) {
      be[i].push_back(j);
      be[j].push_back(i);
    } else {
      ge[i].push_back(j);
      ge[j].push_back(i);
    }
  }
  void bfill(int i, vector<int>& vst) {
    bc[i] = -1;
    vst.push_back(i);
    for (auto& u : be[i]) {
      if (!bc[u]) bfill(u, vst);
    }
  }
  void gfill(int i, vector<int>& vst) {
    gc[i] = -1;
    vst.push_back(i);
    for (auto& u : ge[i]) {
      if (!gc[u]) gfill(u, vst);
    }
  }
  int cmpsize(int i, int col) {
    if (col == 0) {
      if (bc[i] == 0) {
        vector<int> vst;
        bfill(i, vst);
        for (auto& x : vst) {
          bc[x] = vst.size();
        }
      }
      return bc[i];
    } else {
      if (gc[i] == 0) {
        vector<int> vst;
        gfill(i, vst);
        for (auto& x : vst) {
          gc[x] = vst.size();
        }
      }
      return gc[i];
    }
  }
};
int main() {
  int N;
  scanf("%d", &N);
  Tree g(N);
  for (int i = 0; i < N - 1; ++i) {
    int u, v, c;
    scanf("%d%d%d", &u, &v, &c);
    g.add(u - 1, v - 1, c);
  }
  long long ans = 0;
  for (int i = 0; i < N; ++i) {
    ans += (long long)g.cmpsize(i, 0) * g.cmpsize(i, 1) - 1;
  }
  printf("%lld\n", ans);
  return 0;
}
