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<vector<pair<int, int> > > mv;
int val[200001][3];
long long res;
void fi(int ac, int fa) {
  int i, s = mv[ac].size();
  val[ac][0] = val[ac][1] = val[ac][2] = 0;
  for (i = 0; i < s; i++)
    if (mv[ac][i].first != fa) {
      fi(mv[ac][i].first, ac);
      if (mv[ac][i].second) {
        val[ac][1] += val[mv[ac][i].first][1] + 1;
        val[ac][2] += val[mv[ac][i].first][2] + val[mv[ac][i].first][0];
      } else {
        val[ac][0] += val[mv[ac][i].first][0] + 1;
      }
    }
}
void se(int ac, int fa, int ed) {
  if (ed) {
    val[ac][1] = val[fa][1];
    val[ac][2] = val[fa][2] - val[ac][0] + val[fa][0];
  } else
    val[ac][0] = val[fa][0];
  res += val[ac][0] + val[ac][1] + val[ac][2];
  int i, s = mv[ac].size();
  for (i = 0; i < s; i++)
    if (mv[ac][i].first != fa) {
      se(mv[ac][i].first, ac, mv[ac][i].second);
    }
}
int main() {
  int n;
  scanf("%d", &n);
  mv.resize(n);
  int i, u, v, c;
  for (i = 0; i < n - 1; i++) {
    scanf("%d %d %d", &u, &v, &c);
    mv[u - 1].push_back(pair<int, int>(v - 1, c));
    mv[v - 1].push_back(pair<int, int>(u - 1, c));
  }
  fi(0, -1);
  res = val[0][0] + val[0][1] + val[0][2];
  int s = mv[0].size();
  for (i = 0; i < s; i++) se(mv[0][i].first, 0, mv[0][i].second);
  printf("%lld\n", res);
  return 0;
}
