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>
#pragma comment(linker, "/stack:247474112")
#pragma GCC optimize("Ofast")
using namespace std;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
void accelerate() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
const int MAXN = 2e5 + 10;
long long dp[MAXN][4];
long long ans;
int n;
vector<int> E[MAXN];
map<pair<int, int>, int> C;
void DFS(int v = 0, int fa = -1) {
  for (auto u : E[v]) {
    if (u != fa) {
      DFS(u, v);
      if (C[make_pair(min(u, v), max(u, v))] == 0) {
        dp[v][0] += dp[u][0] + 1;
        dp[v][2] += dp[u][2] + dp[u][1];
      } else {
        dp[v][1] += dp[u][1] + 1;
        dp[v][3] += dp[u][3] + dp[u][0];
      }
    }
  }
  ans += 2 * dp[v][0] + 2 * dp[v][1] + dp[v][2] + dp[v][3];
  for (auto u : E[v]) {
    if (u != fa) {
      if (C[make_pair(min(u, v), max(u, v))] == 0) {
        ans += (dp[u][0] + 1) * (dp[v][0] - dp[u][0] - 1);
        ans += (dp[u][0] + 1) * (dp[v][2] - (dp[u][2] + dp[u][1]) + dp[v][1]);
      } else {
        ans += (dp[u][1] + 1) * (dp[v][1] - dp[u][1] - 1);
        ans += (dp[u][0] + dp[u][3]) * (dp[v][1] - dp[u][1] - 1);
      }
    }
  }
}
int main() {
  accelerate();
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    E[x].push_back(y);
    E[y].push_back(x);
    cin >> C[make_pair(min(x, y), max(x, y))];
  }
  DFS();
  cout << ans << endl;
}
