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 MAXN = 200009;
vector<pair<long long, long long> > g[MAXN];
long long n;
long long dp[MAXN][4];
long long ans;
void dfs(long long v, long long pr) {
  for (auto to : g[v]) {
    if (to.first == pr) continue;
    dfs(to.first, v);
    long long cur[4] = {};
    if (to.second == 0) {
      cur[0] = dp[to.first][0] + 1;
      cur[1] = dp[to.first][1] + dp[to.first][3];
    } else {
      cur[3] = dp[to.first][3] + 1;
      cur[2] = dp[to.first][2] + dp[to.first][0];
    }
    ans -= cur[0] * (cur[1] + cur[3]) + cur[2] * cur[3] + cur[0] * cur[0] +
           cur[3] * cur[3];
    for (long long j = 0; j < 4; j++) dp[v][j] += cur[j];
  }
  ans += dp[v][0] * (dp[v][1] + dp[v][3]) + dp[v][2] * dp[v][3] + dp[v][1] +
         dp[v][2] + 2 * (dp[v][0] + dp[v][3]) + dp[v][0] * dp[v][0] +
         dp[v][3] * dp[v][3];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    g[a].emplace_back(b, c);
    g[b].emplace_back(a, c);
  }
  dfs(0, -1);
  cout << ans;
}
