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 = 200000 + 10;
vector<int> v;
int father[maxn][2];
int rk[maxn][2];
int sz[maxn][2];
int Find(int v, int t) {
  return father[v][t] = father[v][t] == v ? v : Find(father[v][t], t);
}
void merge(int x, int y, int t) {
  int a = Find(x, t), b = Find(y, t);
  if (sz[a][t] == 0) sz[a][t]++;
  if (sz[b][t] == 0) sz[b][t]++;
  if (rk[a][t] < rk[b][t]) {
    father[a][t] = b;
    sz[b][t] += sz[a][t];
  } else {
    father[b][t] = a;
    sz[a][t] += sz[b][t];
    if (rk[b][t] == rk[a][t]) {
      ++rk[a][t];
    }
  }
}
int main() {
  int n;
  int u, v, t;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) {
      father[i][0] = father[i][1] = i;
    }
    memset(rk, 0, sizeof(rk));
    memset(sz, 0, sizeof(sz));
    for (int i = 1; i < n; i++) {
      scanf("%d%d%d", &u, &v, &t);
      merge(u, v, t);
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      int a = Find(i, 0), b = Find(i, 1);
      if (sz[a][0] != 0 && sz[b][1] != 0) {
        ans += (long long)(sz[a][0] - 1) * (sz[b][1] - 1);
      }
      if (sz[a][0] != 0) {
        ans += sz[a][0] - 1;
      }
      if (sz[b][1] != 0) {
        ans += sz[b][1] - 1;
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
