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;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "URDL";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 1000003) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000003) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
vector<pair<int, int> > mp[200135];
int sz[2][200135], col[2][200135], cnt;
long long ans;
void dfs(int x, int pa, int w) {
  col[w][x] = cnt;
  sz[w][cnt]++;
  for (auto cp : mp[x])
    if (cp.first != pa && cp.second == w) {
      if (col[w][cp.first] == 0) dfs(cp.first, x, w);
    }
}
void fmain(int tid) {
  scanf("%d", &n);
  for (int(i) = 1; (i) <= (int)(n - 1); (i)++) {
    int u, v, c;
    scanf("%d%d%d", &u, &v, &c);
    mp[u].push_back({v, c});
    mp[v].push_back({u, c});
  }
  for (int(w) = 0; (w) < (int)(2); (w)++) {
    cnt = 0;
    for (int(i) = 1; (i) <= (int)(n); (i)++)
      if (col[w][i] == 0) {
        cnt++;
        dfs(i, 0, w);
        ans += (long long)sz[w][cnt] * (sz[w][cnt] - 1) / 2;
      }
  }
  ans *= 2;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    ans += (long long)(sz[0][col[0][i]] - 1) * (sz[1][col[1][i]] - 1);
  }
  printf("%lld\n", ans);
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}
