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 = 2e5 + 5;
int par[maxn << 1], rankk[maxn << 1], a[maxn << 1], b[maxn];
void init(int n) {
  for (int i = 1; i <= n; i++) par[i] = i, rankk[i] = 0, a[i] = 1;
}
int find(int x) { return par[x] == x ? x : par[x] = find(par[x]); }
void unit(int x, int y) {
  x = find(x), y = find(y);
  if (rankk[x] > rankk[y])
    swap(x, y);
  else if (rankk[x] == rankk[y])
    rankk[y]++;
  par[x] = y;
  a[x] = a[y] = a[x] + a[y];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  init(n << 1);
  for (int i = 1; i <= n - 1; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    if (!c)
      unit(a, b);
    else
      unit(a + n, b + n);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    b[i] = a[find(i)];
    if (b[i] > 1 && find(i) == i) ans += 1ll * b[i] * (b[i] - 1);
  }
  for (int i = 1; i <= n; i++) {
    ans += 1ll * b[i] * (a[find(i + n)] - 1);
  }
  cout << ans << '\n';
  return 0;
}
