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>
const int N = 2e5 + 5;
int fa[N], fb[N];
int da[N], db[N];
int find1(int x) {
  if (x != fa[x]) fa[x] = find1(fa[x]);
  return fa[x];
}
int find2(int x) {
  if (x != fb[x]) fb[x] = find2(fb[x]);
  return fb[x];
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    fa[i] = i, fb[i] = i;
    da[i] = 1, db[i] = 1;
  }
  int x, y, w;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d%d", &x, &y, &w);
    if (w == 1) {
      x = find1(fa[x]), y = find1(fa[y]);
      if (x != y) fa[x] = y;
      da[y] += da[x];
    } else {
      x = find2(fb[x]), y = find2(fb[y]);
      if (x != y) fb[x] = y;
      db[y] += db[x];
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans += (long long)da[find1(i)] * db[find2(i)] - 1;
  }
  printf("%lld\n", ans);
  return 0;
}
