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 int N = 200010;
vector<pair<int, int> > G[N];
long long res = 0LL, dp[N][4];
void dfs(int u, int parent) {
  for (int i = 0; i < G[u].size(); ++i) {
    int v = G[u][i].first;
    int w = G[u][i].second;
    if (v == parent) continue;
    dfs(v, u);
    long long cnt[4] = {0};
    if (w == 0) {
      cnt[0] += dp[v][0] + 1;
      cnt[2] += dp[v][2] + dp[v][3];
    } else {
      cnt[1] += dp[v][0] + dp[v][1];
      cnt[3] += dp[v][3] + 1;
    }
    res += (cnt[0] + cnt[3]) * 2 + cnt[1] + cnt[2];
    res += dp[u][0] * cnt[0] + dp[u][0] * cnt[2] + dp[u][0] * cnt[3];
    res += dp[u][1] * cnt[3];
    res += dp[u][3] * cnt[3];
    res += cnt[0] * dp[u][0] + cnt[0] * dp[u][2] + cnt[0] * dp[u][3];
    res += cnt[1] * dp[u][3];
    res += cnt[3] * dp[u][3];
    for (int j = 0; j < 4; ++j) {
      dp[u][j] += cnt[j];
    }
  }
}
int main() {
  int n;
  cin >> n;
  int a, b, w;
  for (int i = 0; i < n - 1; ++i) {
    cin >> a >> b >> w;
    --a;
    --b;
    G[a].push_back(make_pair(b, w));
    G[b].push_back(make_pair(a, w));
  }
  memset(dp, 0, sizeof(dp));
  dfs(0, -1);
  cout << res << endl;
  return 0;
}
