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 + 10;
const int base = 31337;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int logo = 20;
const int off = 1 << logo;
const int treesiz = off << 1;
int n;
vector<int> graph[maxn][2];
bool bio[maxn];
int col[maxn][2];
int siz[maxn][2];
int dfs(int x, int cl, int b) {
  bio[x] = true;
  col[x][b] = cl;
  int out = 1;
  for (int i = 0; i < graph[x][b].size(); i++) {
    int tren = graph[x][b][i];
    if (!bio[tren]) out += dfs(tren, cl, b);
  }
  return out;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    graph[a][c].push_back(b);
    graph[b][c].push_back(a);
  }
  for (int a = 0; a < 2; a++) {
    memset(bio, false, sizeof bio);
    for (int i = 1; i <= n; i++) {
      if (!bio[i]) {
        siz[i][a] = dfs(i, i, a);
      }
    }
  }
  long long sol = 0;
  for (int i = 1; i <= n; i++) {
    int px = col[i][0];
    int py = col[i][1];
    sol += (long long)siz[px][0] * siz[py][1];
  }
  printf("%lld\n", sol - n);
  return 0;
}
