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 N = 300001;
vector<long long> v1[N], v2[N];
long long c1[N], c2[N], sz[3 * N];
long long col;
long long n;
long long s;
void dfs1(long long i) {
  for (long long j : v1[i]) {
    if (c1[j] == 0) {
      s++;
      c1[j] = col;
      dfs1(j);
    }
  }
}
void dfs2(long long i) {
  for (long long j : v2[i]) {
    if (c2[j] == 0) {
      s++;
      c2[j] = col;
      dfs2(j);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cin >> n;
  memset(c1, 0, sizeof(c1));
  memset(c2, 0, sizeof(c2));
  memset(sz, 0, sizeof(sz));
  col = 1;
  for (long long i = 0; i < n - 1; i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    if (c == 0) {
      v1[a].push_back(b);
      v1[b].push_back(a);
    } else {
      v2[a].push_back(b);
      v2[b].push_back(a);
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    if (c1[i] == 0) {
      s = 1;
      c1[i] = col;
      dfs1(i);
      sz[col] = s;
      ans += (s * (s - 1));
      col++;
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (c2[i] == 0) {
      s = 1;
      c2[i] = col;
      dfs2(i);
      sz[col] = s;
      ans += (s * (s - 1));
      col++;
    }
  }
  for (long long i = 1; i <= n; i++) {
    long long x = sz[c1[i]];
    long long y = sz[c2[i]];
    ans += (x - 1) * (y - 1);
  }
  cout << ans;
}
