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 inf = 1e9 + 7;
const double eps = 1e-10;
const long long linf = 1e18 + 7;
const long long hh = 523;
int dp[2][200007];
int ans[2][200007];
vector<int> G[2][200007];
int fa[200007];
long long res = 0, sum0 = 0, sum1 = 0;
void dfs(int v, int p) {
  fa[v] = p;
  dp[0][v] = 1;
  dp[1][v] = 0;
  for (int k = 0; k <= 1; k++) {
    for (int j = 0; j < G[k][v].size(); j++) {
      int u = G[k][v][j];
      if (u == p) {
        continue;
      }
      dfs(u, v);
      if (k == 0) {
        dp[0][v] += dp[0][u];
      } else {
        dp[0][v] += dp[1][u] + 1;
        dp[1][v] += dp[1][u] + 1;
      }
    }
  }
}
void calc(int v, int e) {
  if (v == 0) {
    res += dp[0][0];
  } else {
    if (e == 1) {
      res += dp[0][v] + sum1;
    } else {
      res += dp[0][v] + sum0;
    }
  }
  for (int k = 0; k <= 1; k++) {
    for (int j = 0; j < G[k][v].size(); j++) {
      int u = G[k][v][j];
      if (u == fa[v]) {
        continue;
      }
      long long t0 = sum0, t1 = sum1;
      if (k == 1) {
        sum1 += dp[1][v] - dp[1][u];
        sum0 = sum1;
      } else {
        sum0 += dp[0][v] - dp[0][u];
        sum1 = 0;
      }
      calc(u, k);
      sum0 = t0;
      sum1 = t1;
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int x, y, c;
    scanf("%d%d%d", &x, &y, &c);
    x--;
    y--;
    G[c][x].push_back(y);
    G[c][y].push_back(x);
  }
  dfs(0, 0);
  calc(0, 0);
  printf("%I64d\n", res - n);
  return 0;
}
