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 + 7;
int far[maxn][2];
int cnt[maxn][2];
void init() {
  for (int i = 0; i < maxn; i++)
    far[i][0] = far[i][1] = i, cnt[i][0] = cnt[i][1] = 1;
}
int Find(int x, int type) {
  return far[x][type] == x ? x : far[x][type] = Find(far[x][type], type);
}
void Union(int x, int y, int type) {
  x = Find(x, type), y = Find(y, type);
  if (x == y) return;
  cnt[x][type] += cnt[y][type];
  far[y][type] = far[x][type];
}
int main() {
  init();
  int n;
  long long ans = 0;
  scanf("%d", &n);
  for (int i = 1, u, v, ty; i < n; i++) {
    scanf("%d%d%d", &u, &v, &ty);
    Union(u, v, ty);
  }
  for (int i = 1; i <= n; i++) {
    int far0 = Find(i, 0), far1 = Find(i, 1);
    if (far0 == i) ans += 1LL * cnt[i][0] * (cnt[i][0] - 1);
    if (far1 == i) ans += 1LL * cnt[i][1] * (cnt[i][1] - 1);
    ans += 1LL * (cnt[far0][0] - 1) * (cnt[far1][1] - 1);
  }
  printf("%lld\n", ans);
}
