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;
inline long long read() {
  char c;
  long long sign = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') sign = -1;
  long long res = (c - '0');
  while ((c = getchar()) >= '0' && c <= '9')
    res = (res << 1) + (res << 3) + c - '0';
  return res * sign;
}
inline void print(long long x) {
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  print(x);
}
const long long maxn = 400005;
long long c1, c2, n1, n2, n, m;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long ans;
long long f0[maxn], f1[maxn];
long long size0[maxn], size1[maxn];
long long find0(long long x) { return x == f0[x] ? x : f0[x] = find0(f0[x]); }
long long find1(long long x) { return x == f1[x] ? x : f1[x] = find1(f1[x]); }
void add(long long x, long long y, long long z) {
  if (z == 1) {
    long long fx = find1(x), fy = find1(y);
    f1[fx] = fy;
    size1[fy] += size1[fx];
  } else if (z == 0) {
    long long fx = find0(x), fy = find0(y);
    f0[fx] = fy;
    size0[fy] += size0[fx];
  }
}
void solve() {
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    long long k1 = find0(i), k2 = find1(i);
    ans += (1ll * size0[k1] * size1[k2] - 1);
  }
  write(ans);
}
signed main() {
  n = read();
  long long x, y, z;
  for (long long i = 1; i <= n; i++) f0[i] = i, f1[i] = i;
  for (long long i = 1; i <= n; i++) size0[i] = size1[i] = 1;
  for (long long i = 1; i <= n - 1; i++) {
    x = read(), y = read(), z = read();
    add(x, y, z);
  }
  solve();
}
