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 + 100;
struct Node {
  int from, to, value;
} edge[maxN + 1];
int pa[maxN + 1], sz0[maxN + 1], sz1[maxN + 1];
int n;
long long ans;
inline int read() {
  int num = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) num = (num << 3) + (num << 1) + (ch ^ 48), ch = getchar();
  return num * f;
}
inline int find(int x) { return pa[x] == x ? x : pa[x] = find(pa[x]); }
int main() {
  n = read();
  for (int i = 1; i < n; i++)
    edge[i].from = read(), edge[i].to = read(), edge[i].value = read();
  for (int i = 1; i <= n; i++) pa[i] = i, sz0[i] = 1;
  for (int i = 1; i < n; i++)
    if (edge[i].value) {
      int x = edge[i].from, y = edge[i].to;
      x = find(x), y = find(y);
      if (x == y) continue;
      pa[x] = y;
      sz0[y] += sz0[x];
    }
  for (int i = 1; i <= n; i++)
    if (find(i) == i) ans += 1ll * sz0[i] * (sz0[i] - 1);
  for (int i = 1; i <= n; i++) sz0[i] = sz0[find(i)];
  for (int i = 1; i <= n; i++) pa[i] = i, sz1[i] = 1;
  for (int i = 1; i < n; i++)
    if (!edge[i].value) {
      int x = edge[i].from, y = edge[i].to;
      x = find(x), y = find(y);
      if (x == y) continue;
      pa[x] = y;
      sz1[y] += sz1[x];
    }
  for (int i = 1; i <= n; i++)
    if (find(i) == i) ans += 1ll * sz1[i] * (sz1[i] - 1);
  for (int i = 1; i <= n; i++) sz1[i] = sz1[find(i)];
  for (int i = 1; i <= n; i++) ans += 1ll * (sz0[i] - 1) * (sz1[i] - 1);
  printf("%lld", ans);
  return 0;
}
