#include <bits/stdc++.h>
int n, fa[200005][2], sz[200005][2];
long long ans;
int find(int x, int c) {
  return fa[x][c] == x ? x : (fa[x][c] = find(fa[x][c], c));
}
void merge(int x, int y, int c) {
  x = find(x, c), y = find(y, c);
  if (x == y) return;
  fa[y][c] = fa[x][c], sz[x][c] += sz[y][c];
}
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i)
    fa[i][0] = fa[i][1] = i, sz[i][0] = sz[i][1] = 1;
  for (register int i = 1, u, v, w; i < n; ++i)
    scanf("%d%d%d", &u, &v, &w), merge(u, v, w);
  for (register int i = 1; i <= n; ++i)
    ans += 1ll * sz[find(i, 0)][0] * sz[find(i, 1)][1] - 1;
  printf("%lld", ans);
}

### Prompt

Generate a Cpp solution to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
int n, fa[200005][2], sz[200005][2];
long long ans;
int find(int x, int c) {
  return fa[x][c] == x ? x : (fa[x][c] = find(fa[x][c], c));
}
void merge(int x, int y, int c) {
  x = find(x, c), y = find(y, c);
  if (x == y) return;
  fa[y][c] = fa[x][c], sz[x][c] += sz[y][c];
}
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i)
    fa[i][0] = fa[i][1] = i, sz[i][0] = sz[i][1] = 1;
  for (register int i = 1, u, v, w; i < n; ++i)
    scanf("%d%d%d", &u, &v, &w), merge(u, v, w);
  for (register int i = 1; i <= n; ++i)
    ans += 1ll * sz[find(i, 0)][0] * sz[find(i, 1)][1] - 1;
  printf("%lld", ans);
}

```