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 inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long _INF = 0xc0c0c0c0c0c0c0c0;
const long long mod = (int)1e9 + 7;
const int N = 2e5 + 100;
vector<pair<int, int> > vc[N];
int dp[N][2];
void dfs1(int o, int u) {
  dp[u][0] = dp[u][1] = 0;
  for (pair<int, int>& t : vc[u]) {
    if (o == t.first) continue;
    dfs1(u, t.first);
    if (t.second == 0)
      dp[u][0] += dp[t.first][0] + 1;
    else
      dp[u][1] += dp[t.first][0] + dp[t.first][1] + 1;
  }
}
long long ans = 0;
void dfs2(int o, int u, int f) {
  if (o) {
    int t[2] = {dp[o][0], dp[o][1]};
    if (f == 0)
      t[0] -= dp[u][0] + 1;
    else
      t[1] -= dp[u][0] + dp[u][1] + 1;
    if (f == 0)
      dp[u][0] += t[0] + 1;
    else
      dp[u][1] += t[0] + t[1] + 1;
  }
  ans += dp[u][0] + dp[u][1];
  for (pair<int, int>& t : vc[u]) {
    if (o == t.first) continue;
    dfs2(u, t.first, t.second);
  }
}
int main() {
  int n, x, y, v;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d%d", &x, &y, &v);
    vc[x].push_back(pair<int, int>(y, v));
    vc[y].push_back(pair<int, int>(x, v));
  }
  dfs1(0, 1);
  dfs2(0, 1, 0);
  cout << ans << endl;
  return 0;
}
