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 + 5;
const int mod = 998244353;
long long qp(long long a, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
template <class T>
inline bool scan(T& ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
template <class T>
inline void out(T x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
int n;
int sz[maxn], szz[maxn];
int pre[maxn], ppre[maxn];
int F(int x) { return x == pre[x] ? x : pre[x] = F(pre[x]); }
int Find(int x) { return x == ppre[x] ? x : ppre[x] = Find(ppre[x]); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    sz[i] = szz[i] = 1;
    pre[i] = ppre[i] = i;
  }
  for (int i = 1, x, y, w; i < n; ++i) {
    cin >> x >> y >> w;
    if (w == 0) {
      int u = F(x);
      int v = F(y);
      if (u == v) continue;
      sz[u] += sz[v];
      pre[v] = u;
    } else {
      int u = Find(x);
      int v = Find(y);
      if (u == v) continue;
      szz[u] += szz[v];
      ppre[v] = u;
    }
  }
  long long res = 0;
  for (int i = 1; i <= n; ++i) {
    res += 1ll * sz[F(i)] * szz[Find(i)];
  }
  res -= n;
  cout << res << endl;
  return 0;
}
