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 long long NR = 2e5 + 100, oo = 1e9 + 10, MOD = 1e9 + 7;
const long double pi = 2 * acos(0.0);
vector<pair<int, int> > v[NR];
long long n, dp0[NR], dp1[NR];
long long sol;
void dfs1(int nod, int fat) {
  int vecin, cost;
  vector<pair<int, int> >::iterator it;
  dp0[nod] = 1;
  dp1[nod] = 1;
  for (it = v[nod].begin(); it != v[nod].end(); ++it) {
    vecin = (*it).first;
    cost = (*it).second;
    if (vecin == fat) continue;
    dfs1(vecin, nod);
    if (cost == 1) {
      dp1[nod] += dp1[vecin];
      dp0[nod] += dp1[vecin];
    } else {
      dp0[nod] += dp0[vecin];
    }
  }
}
void dfs2(int nod, int fat, int tip, long long sum) {
  int vecin, cost;
  vector<pair<int, int> >::iterator it;
  if (tip == 1) {
    dp0[nod] += sum;
    dp1[nod] += sum;
  }
  if (tip == 0) {
    dp0[nod] += sum;
  }
  for (it = v[nod].begin(); it != v[nod].end(); ++it) {
    vecin = (*it).first;
    cost = (*it).second;
    if (vecin == fat) continue;
    if (cost == 1) {
      dfs2(vecin, nod, 1, dp1[nod] - dp1[vecin]);
    }
    if (cost == 0) {
      dfs2(vecin, nod, 0, dp0[nod] - dp0[vecin]);
    }
  }
  sol += dp0[nod] - 1;
}
int main() {
  int x, y, z, i;
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (i = 1; i < n; ++i) {
    cin >> x >> y >> z;
    v[x].push_back(make_pair(y, z));
    v[y].push_back(make_pair(x, z));
  }
  dfs1(1, 0);
  dfs2(1, 0, -1, 0);
  cout << sol << '\n';
  return 0;
}
