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 N = 2e5 + 5;
int read() {
  int s = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    s = s * 10 + c - '0';
    c = getchar();
  }
  return s * f;
}
int n;
long long Ans;
struct Graph {
  int fa[N], sum[N];
  void Init() {
    for (int i = 1; i <= n; i++) {
      sum[i] = 1;
      fa[i] = i;
    }
  }
  int Find(int x) { return x == fa[x] ? x : fa[x] = Find(fa[x]); }
  void Link(int x, int y) {
    x = Find(x);
    y = Find(y);
    if (x != y) {
      fa[x] = y;
      sum[y] += sum[x];
    }
  }
  int Sum(int x) { return sum[Find(x)]; }
} G[2];
int main() {
  n = read();
  G[0].Init();
  G[1].Init();
  for (int i = 1; i < n; i++) {
    int x = read(), y = read(), op = read();
    G[op].Link(x, y);
  }
  for (int i = 1; i <= n; i++) Ans += 1ll * G[0].Sum(i) * G[1].Sum(i);
  printf("%lld\n", Ans - n);
  return 0;
}
