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 maxn = 4e5 + 5;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
long long co[maxn];
long long n, m, k, t;
string str;
vector<pair<int, int> > ff;
long long f[2][maxn], cnt[2][maxn];
void init(int n) {
  for (int i = 1; i <= n; i++) f[0][i] = f[1][i] = i, cnt[0][i] = cnt[1][i] = 1;
}
int Find(int x, int flag) {
  return f[flag][x] == x ? x : f[flag][x] = Find(f[flag][x], flag);
}
void Union(int x, int y, int flag) {
  int fx = Find(x, flag), fy = Find(y, flag);
  f[flag][fx] = fy;
  cnt[flag][fy] += cnt[flag][fx];
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  init(n);
  for (int i = 1; i <= n - 1; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    Union(a, b, c);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (f[0][i] == i) ans += cnt[0][i] * (cnt[0][i] - 1);
    if (f[1][i] == i) ans += cnt[1][i] * (cnt[1][i] - 1);
    ans += (cnt[0][Find(i, 0)] - 1) * (cnt[1][Find(i, 1)] - 1);
  }
  cout << ans;
}
