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;
int n;
int fa[2][200000 + 1109];
long long s[2][200000 + 1109];
long long Sum[200000 + 1109];
long long Ans = 0;
int find(int x, int z) {
  if (fa[z][x] == x) return x;
  fa[z][x] = find(fa[z][x], z);
  return fa[z][x];
}
void Union(int x, int y, int z) {
  int fx = find(x, z);
  int fy = find(y, z);
  if (fx == fy) return;
  fa[z][fy] = fx;
  s[z][fx] += s[z][fy];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) fa[0][i] = fa[1][i] = i, s[0][i] = s[1][i] = 1;
  int x, y, z;
  for (int i = 1; i < n; i++) {
    scanf("%d%d%d", &x, &y, &z);
    Union(x, y, z);
  }
  int f0, f1;
  for (int i = 1; i <= n; i++) {
    f0 = find(i, 0);
    f1 = find(i, 1);
    if (s[0][f0] == 1 || s[1][f1] == 1) continue;
    Sum[f1] += s[0][f0] - 1;
    Ans -= s[0][f0] - 1;
  }
  for (int i = 1; i <= n; i++) {
    if (fa[1][i] == i) Ans += (Sum[i] + s[1][i] - 1) * s[1][i];
    if (fa[0][i] == i) Ans += s[0][i] * (s[0][i] - 1);
  }
  printf("%I64d\n", Ans);
}
