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;
struct st1 {
  long long n, v;
};
long long a, b, c, N, M, t, ans, fa0[200007], fa1[200007];
long long n0[200007], n1[200007], n2[200007];
int get(long long *fa, long long x1) {
  if (fa[x1] == x1) return fa[x1];
  fa[x1] = get(fa, fa[x1]);
  return fa[x1];
}
int main() {
  scanf("%I64d", &N);
  for (int i = 1; i <= N; i++) fa0[i] = i, fa1[i] = i, n0[i] = 1, n1[i] = 1;
  for (int i = 1; i <= N - 1; i++) {
    scanf("%I64d%I64d%I64d", &a, &b, &c);
    if (!c) {
      a = get(fa0, a);
      b = get(fa0, b);
      if (a != b) {
        fa0[a] = b;
        n0[b] += n0[a];
      }
    } else {
      a = get(fa1, a);
      b = get(fa1, b);
      if (a != b) {
        fa1[a] = b;
        n1[b] += n1[a];
      }
    }
  }
  for (int i = 1; i <= N; i++) n2[get(fa0, i)] += n1[get(fa1, i)];
  for (int i = 1; i <= N; i++) ans += n2[get(fa0, i)] - 1;
  printf("%I64d\n", ans);
  return 0;
}
