#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int N = 2e5 + 5;
int n, m;
int pre[2][N], cnt[2][N];
int Find(int x, int d) {
  return pre[d][x] == x ? x : pre[d][x] = Find(pre[d][x], d);
}
void join(int x, int y, int d) {
  int fx = Find(x, d), fy = Find(y, d);
  if (fx != fy) {
    pre[d][fx] = fy;
    cnt[d][fy] += cnt[d][fx];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i <= n; i++)
    pre[0][i] = pre[1][i] = i, cnt[0][i] = cnt[1][i] = 1;
  for (int i = 1; i <= n - 1; i++) {
    int x, y, d;
    scanf("%d%d%d", &x, &y, &d);
    join(x, y, d);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int fx = Find(i, 0), fy = Find(i, 1);
    ans += 1ll * cnt[0][fx] * cnt[1][fy] - 1;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int N = 2e5 + 5;
int n, m;
int pre[2][N], cnt[2][N];
int Find(int x, int d) {
  return pre[d][x] == x ? x : pre[d][x] = Find(pre[d][x], d);
}
void join(int x, int y, int d) {
  int fx = Find(x, d), fy = Find(y, d);
  if (fx != fy) {
    pre[d][fx] = fy;
    cnt[d][fy] += cnt[d][fx];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i <= n; i++)
    pre[0][i] = pre[1][i] = i, cnt[0][i] = cnt[1][i] = 1;
  for (int i = 1; i <= n - 1; i++) {
    int x, y, d;
    scanf("%d%d%d", &x, &y, &d);
    join(x, y, d);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int fx = Find(i, 0), fy = Find(i, 1);
    ans += 1ll * cnt[0][fx] * cnt[1][fy] - 1;
  }
  printf("%lld\n", ans);
  return 0;
}

```