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 P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e6 + 10;
int n, sz1[N], sz2[N], fa1[N], fa2[N];
int f1(int first) { return fa1[first] ? fa1[first] = f1(fa1[first]) : first; }
int f2(int first) { return fa2[first] ? fa2[first] = f2(fa2[first]) : first; }
void add1(int first, int second) {
  if ((first = f1(first)) != (second = f1(second)))
    fa1[first] = second, sz1[second] += sz1[first];
}
void add2(int first, int second) {
  if ((first = f2(first)) != (second = f2(second)))
    fa2[first] = second, sz2[second] += sz2[first];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) sz1[i] = sz2[i] = 1;
  for (int i = 2; i <= n; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    w ? add2(u, v) : add1(u, v);
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) ans += (long long)sz1[f1(i)] * sz2[f2(i)] - 1;
  printf("%lld\n", ans);
}
