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;
int n, u, v, c;
long long int dp[200010][5];
vector<int> edge[200010], col[200010];
long long int ans = 0;
void dfs(int cn, int par = -1) {
  long long tans = ans;
  for (int i = 0; i < edge[cn].size(); i++) {
    int to = edge[cn][i];
    int color = col[cn][i];
    if (to == par) continue;
    dfs(to, cn);
    long long ttans = 0;
    if (color == 0) {
      ttans = 2 * (dp[cn][0] + 1) * (dp[to][0] + 1) +
              (dp[cn][0] + 1) * (dp[to][1] + dp[to][2]) +
              (dp[to][0] + 1) * (dp[cn][1] + dp[cn][2]);
      dp[cn][0] += dp[to][0] + 1;
      dp[cn][2] += dp[to][2] + dp[to][1];
    } else {
      ttans = 2 * (dp[cn][1] + 1) * (dp[to][1] + 1) +
              dp[cn][0] * (dp[to][1] + 1) + (dp[to][0]) * (dp[cn][1] + 1) +
              dp[cn][3] * (dp[to][1] + 1) + (dp[to][3]) * (dp[cn][1] + 1);
      dp[cn][1] += dp[to][1] + 1;
      dp[cn][3] += dp[to][3] + dp[to][0];
    }
    ans += ttans;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> u >> v >> c;
    edge[u].push_back(v);
    edge[v].push_back(u);
    col[u].push_back(c);
    col[v].push_back(c);
  }
  dfs(1);
  cout << ans << endl;
}
